[{"Name":"Introduction to Inner Product Spaces, Normed Spaces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Inner Product Space","Duration":"2m 28s","ChapterTopicVideoID":27445,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.214","Text":"In this clip, we\u0027ll define the concept of an inner product space."},{"Start":"00:05.214 ","End":"00:08.935","Text":"You may have encountered this in linear algebra,"},{"Start":"00:08.935 ","End":"00:11.390","Text":"but we won\u0027t assume anything here,"},{"Start":"00:11.390 ","End":"00:12.665","Text":"we\u0027ll start from scratch."},{"Start":"00:12.665 ","End":"00:16.825","Text":"But we\u0027re only going to consider in the product spaces over 2 fields,"},{"Start":"00:16.825 ","End":"00:19.015","Text":"complex numbers or real numbers."},{"Start":"00:19.015 ","End":"00:21.700","Text":"In general, an inner product space is defined over a field,"},{"Start":"00:21.700 ","End":"00:23.965","Text":"that we\u0027ll just take as 1 of these 2."},{"Start":"00:23.965 ","End":"00:25.824","Text":"Now the definition."},{"Start":"00:25.824 ","End":"00:29.770","Text":"We start with a vector space over a field F,"},{"Start":"00:29.770 ","End":"00:31.120","Text":"which is either, I can say,"},{"Start":"00:31.120 ","End":"00:33.160","Text":"complex or real numbers,"},{"Start":"00:33.160 ","End":"00:36.280","Text":"and we define an inner product,"},{"Start":"00:36.280 ","End":"00:39.610","Text":"which is a function that takes a pair of elements from"},{"Start":"00:39.610 ","End":"00:42.875","Text":"V and gives us a member of the field."},{"Start":"00:42.875 ","End":"00:45.815","Text":"In other words, we take 2 vectors, u and v,"},{"Start":"00:45.815 ","End":"00:50.825","Text":"and they give a scalar in F and we denote it by angular brackets,"},{"Start":"00:50.825 ","End":"00:52.235","Text":"u, v,"},{"Start":"00:52.235 ","End":"00:57.350","Text":"also dot product u.v, either or."},{"Start":"00:57.350 ","End":"01:00.110","Text":"But it has to satisfy certain axioms,"},{"Start":"01:00.110 ","End":"01:02.645","Text":"not just any function like this."},{"Start":"01:02.645 ","End":"01:04.670","Text":"The axioms are, first of all,"},{"Start":"01:04.670 ","End":"01:07.475","Text":"it has to be linear in the first argument."},{"Start":"01:07.475 ","End":"01:10.225","Text":"The first argument means what\u0027s to the left of the comma."},{"Start":"01:10.225 ","End":"01:12.345","Text":"u plus w,"},{"Start":"01:12.345 ","End":"01:13.965","Text":"v is u,"},{"Start":"01:13.965 ","End":"01:15.390","Text":"v plus w,"},{"Start":"01:15.390 ","End":"01:17.970","Text":"v. That\u0027s the additivity."},{"Start":"01:17.970 ","End":"01:19.975","Text":"Then the scalar multiplication,"},{"Start":"01:19.975 ","End":"01:23.030","Text":"Alpha is assumed to be in the field F."},{"Start":"01:23.030 ","End":"01:26.330","Text":"We can multiply a scalar times a vector, Alpha u,"},{"Start":"01:26.330 ","End":"01:29.525","Text":"v comes out to be Alpha times u,"},{"Start":"01:29.525 ","End":"01:35.855","Text":"v. Second axiom is called conjugate symmetry."},{"Start":"01:35.855 ","End":"01:37.609","Text":"If we reverse the order,"},{"Start":"01:37.609 ","End":"01:39.530","Text":"we get the conjugate."},{"Start":"01:39.530 ","End":"01:42.260","Text":"Of course, if it\u0027s over the real numbers,"},{"Start":"01:42.260 ","End":"01:47.375","Text":"then it will be completely symmetric because the conjugate of a real number is itself."},{"Start":"01:47.375 ","End":"01:52.780","Text":"This is also called Hermitian symmetry as mathematician Hermit."},{"Start":"01:52.780 ","End":"01:56.885","Text":"The third axiom is called positive definiteness."},{"Start":"01:56.885 ","End":"02:01.632","Text":"The inner product of a vector with itself is always bigger or equal to 0,"},{"Start":"02:01.632 ","End":"02:07.315","Text":"and it\u0027s actually equal to 0 if and only if the vector is 0."},{"Start":"02:07.315 ","End":"02:09.650","Text":"Finally, we come to inner product space,"},{"Start":"02:09.650 ","End":"02:13.295","Text":"a vector space together with an inner product on it,"},{"Start":"02:13.295 ","End":"02:14.930","Text":"which satisfies these axioms,"},{"Start":"02:14.930 ","End":"02:17.390","Text":"is called an inner product space."},{"Start":"02:17.390 ","End":"02:19.040","Text":"If it\u0027s over the reals,"},{"Start":"02:19.040 ","End":"02:21.170","Text":"it\u0027s called a real inner product space."},{"Start":"02:21.170 ","End":"02:22.550","Text":"If the field is the complex,"},{"Start":"02:22.550 ","End":"02:25.195","Text":"we call it a complex inner product space."},{"Start":"02:25.195 ","End":"02:28.890","Text":"That is it for this introductory clip."}],"ID":28648},{"Watched":false,"Name":"Exercise 1","Duration":"3m 30s","ChapterTopicVideoID":27450,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.600","Text":"In this exercise, v is the space of"},{"Start":"00:03.600 ","End":"00:08.330","Text":"continuously differentiable functions on the interval minus 1 to 1."},{"Start":"00:08.330 ","End":"00:10.615","Text":"This is notation."},{"Start":"00:10.615 ","End":"00:13.199","Text":"For any two functions,"},{"Start":"00:13.199 ","End":"00:14.220","Text":"f and g and v,"},{"Start":"00:14.220 ","End":"00:18.975","Text":"we define the product of f and g, the following expression."},{"Start":"00:18.975 ","End":"00:21.930","Text":"Note that here\u0027s the functions f and g,"},{"Start":"00:21.930 ","End":"00:23.730","Text":"and here\u0027s the derivatives."},{"Start":"00:23.730 ","End":"00:26.720","Text":"We have to prove that this is an inner product on v,"},{"Start":"00:26.720 ","End":"00:31.005","Text":"and here\u0027s a reminder of the axioms for inner product."},{"Start":"00:31.005 ","End":"00:34.200","Text":"Also a reminder of what the product is."},{"Start":"00:34.200 ","End":"00:42.435","Text":"Linearity. The additivity f plus g with h gives us the following expression."},{"Start":"00:42.435 ","End":"00:46.895","Text":"Then we can break it up using distributivity"},{"Start":"00:46.895 ","End":"00:51.350","Text":"of multiplication over addition and also additivity of the integrals."},{"Start":"00:51.350 ","End":"00:53.585","Text":"We can break it up as follows."},{"Start":"00:53.585 ","End":"00:57.905","Text":"This with this give us f with h,"},{"Start":"00:57.905 ","End":"01:00.560","Text":"and this with this give us a product of g with h,"},{"Start":"01:00.560 ","End":"01:02.645","Text":"and that\u0027s what we had to show in this part."},{"Start":"01:02.645 ","End":"01:05.495","Text":"Now there\u0027s multiplication with a scalar."},{"Start":"01:05.495 ","End":"01:09.200","Text":"Here we put Alpha f instead of f,"},{"Start":"01:09.200 ","End":"01:15.750","Text":"and what we can do is take Alpha outside the function and outside the integral,"},{"Start":"01:15.750 ","End":"01:22.470","Text":"and here also Alpha times f prime is Alpha times f prime and the Alpha can come out."},{"Start":"01:22.470 ","End":"01:28.290","Text":"We have this and then just take Alpha outside the brackets, and this,"},{"Start":"01:28.290 ","End":"01:35.015","Text":"and this is just f product with g. That\u0027s the linearity."},{"Start":"01:35.015 ","End":"01:38.990","Text":"Next axiom is conjugate symmetry. Let\u0027s see."},{"Start":"01:38.990 ","End":"01:43.010","Text":"Does this equal this thought with the right-hand side,"},{"Start":"01:43.010 ","End":"01:47.750","Text":"the conjugative of g product with f is the following,"},{"Start":"01:47.750 ","End":"01:53.945","Text":"and then the conjugate of a sum is just the sum of the conjugates."},{"Start":"01:53.945 ","End":"01:56.795","Text":"The conjugate goes inside the integral,"},{"Start":"01:56.795 ","End":"01:59.240","Text":"and then the conjugate of a product is the product"},{"Start":"01:59.240 ","End":"02:01.910","Text":"of the conjugates so we end up with this,"},{"Start":"02:01.910 ","End":"02:07.385","Text":"and then we can just switch the order of the product here and here and get"},{"Start":"02:07.385 ","End":"02:13.460","Text":"exactly f product with g. The next axiom is the positive definite,"},{"Start":"02:13.460 ","End":"02:17.900","Text":"and thus f with f is the following."},{"Start":"02:17.900 ","End":"02:22.630","Text":"That\u0027s the absolute value of f ^ integral,"},{"Start":"02:22.630 ","End":"02:26.120","Text":"and here the absolute value of f prime ^,"},{"Start":"02:26.120 ","End":"02:30.230","Text":"the integral, well, the integrand is non-negative here and here."},{"Start":"02:30.230 ","End":"02:33.485","Text":"So of course we\u0027ll get something bigger or equal to 0."},{"Start":"02:33.485 ","End":"02:36.170","Text":"Suppose this is equal to 0."},{"Start":"02:36.170 ","End":"02:42.050","Text":"This will be true if and only if the following expression is 0."},{"Start":"02:42.050 ","End":"02:44.435","Text":"This will be true."},{"Start":"02:44.435 ","End":"02:47.420","Text":"Both non-negative, we can just take one of them."},{"Start":"02:47.420 ","End":"02:52.440","Text":"It will be true only if the first one is equal to 0."},{"Start":"02:52.440 ","End":"02:57.110","Text":"Actually not an if and only if it\u0027s just a one-way implication."},{"Start":"02:57.110 ","End":"03:02.900","Text":"Here, the integral of a non-negative continuous function is"},{"Start":"03:02.900 ","End":"03:10.403","Text":"0 means that the function is 0 and if the absolute value squared of f is 0,"},{"Start":"03:10.403 ","End":"03:13.250","Text":"that means that f is 0,"},{"Start":"03:13.250 ","End":"03:17.150","Text":"which means that f is 0."},{"Start":"03:17.150 ","End":"03:19.490","Text":"As far as the other direction goes,"},{"Start":"03:19.490 ","End":"03:22.150","Text":"it\u0027s clear that if f is 0,"},{"Start":"03:22.150 ","End":"03:27.470","Text":"then this expression is 0 and so this is 0."},{"Start":"03:27.470 ","End":"03:30.690","Text":"That concludes this exercise."}],"ID":28649},{"Watched":false,"Name":"Exercise 2","Duration":"59s","ChapterTopicVideoID":27455,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.804","Text":"In this exercise, we\u0027ll prove a version,"},{"Start":"00:02.804 ","End":"00:06.990","Text":"variant of Pythagoras\u0027 theorem for inner product spaces,"},{"Start":"00:06.990 ","End":"00:09.300","Text":"and the way we phrase it is like this."},{"Start":"00:09.300 ","End":"00:11.475","Text":"Let v be an inner product space,"},{"Start":"00:11.475 ","End":"00:17.115","Text":"and let u and v be such that the inner product of u and v is 0."},{"Start":"00:17.115 ","End":"00:23.310","Text":"This is also phrased as u is perpendicular to v. We have to prove that"},{"Start":"00:23.310 ","End":"00:29.805","Text":"the norm of (u+v)^2 equals the norm of u^2 plus the norm of v^2, and here it goes."},{"Start":"00:29.805 ","End":"00:33.680","Text":"Norm of (u+v)^2 is u plus v dot product with u plus"},{"Start":"00:33.680 ","End":"00:38.075","Text":"v. We\u0027ve seen this many times before and we expanded like so."},{"Start":"00:38.075 ","End":"00:42.290","Text":"Then, because u dot product with v is 0,"},{"Start":"00:42.290 ","End":"00:43.840","Text":"this is 0,"},{"Start":"00:43.840 ","End":"00:46.640","Text":"and this is the conjugate of this,"},{"Start":"00:46.640 ","End":"00:49.190","Text":"but the conjugate of 0 is also 0."},{"Start":"00:49.190 ","End":"00:52.400","Text":"So this and this is 0 and we\u0027re just left with these two and"},{"Start":"00:52.400 ","End":"00:56.280","Text":"that\u0027s norm of u^2 and that\u0027s norm of v^2,"},{"Start":"00:56.280 ","End":"00:59.410","Text":"and that concludes the proof."}],"ID":28650},{"Watched":false,"Name":"Exercise 3","Duration":"1m 51s","ChapterTopicVideoID":27456,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.850","Text":"In this exercise, V is the space of"},{"Start":"00:02.850 ","End":"00:09.190","Text":"twice continuously differentiable real functions on the interval a, b."},{"Start":"00:09.230 ","End":"00:13.335","Text":"What it means is that f is in this space,"},{"Start":"00:13.335 ","End":"00:18.090","Text":"if f has a second derivative which is continuous on a b."},{"Start":"00:18.090 ","End":"00:20.925","Text":"This is the notation for this space."},{"Start":"00:20.925 ","End":"00:26.190","Text":"Now, we\u0027ll define a product like so and the question is,"},{"Start":"00:26.190 ","End":"00:28.410","Text":"is this an inner product on V?"},{"Start":"00:28.410 ","End":"00:31.005","Text":"Does it satisfy all the axioms?"},{"Start":"00:31.005 ","End":"00:33.885","Text":"It turns out that the answer is no."},{"Start":"00:33.885 ","End":"00:38.735","Text":"There\u0027s one axiom it doesn\u0027t satisfy the one that says that"},{"Start":"00:38.735 ","End":"00:44.680","Text":"the inner product of a vector with itself is 0 if and only if that vector is 0."},{"Start":"00:44.680 ","End":"00:47.150","Text":"Even this works one way."},{"Start":"00:47.150 ","End":"00:50.975","Text":"If f is 0, certainly f is 0."},{"Start":"00:50.975 ","End":"00:52.565","Text":"But the other way round,"},{"Start":"00:52.565 ","End":"00:54.800","Text":"if the inner product of f with itself is 0,"},{"Start":"00:54.800 ","End":"00:56.840","Text":"does not imply that f is 0."},{"Start":"00:56.840 ","End":"00:58.220","Text":"May 0 or may not be,"},{"Start":"00:58.220 ","End":"01:00.890","Text":"but not necessarily. Let\u0027s see."},{"Start":"01:00.890 ","End":"01:03.545","Text":"If this inner product is 0,"},{"Start":"01:03.545 ","End":"01:07.850","Text":"then the integral of the second derivative of f"},{"Start":"01:07.850 ","End":"01:14.884","Text":"squared is 0 and this means that as we\u0027ve seen before,"},{"Start":"01:14.884 ","End":"01:20.360","Text":"f double-prime squared is 0 everywhere because it\u0027s continuous,"},{"Start":"01:20.360 ","End":"01:23.525","Text":"and then this is 0 everywhere."},{"Start":"01:23.525 ","End":"01:29.120","Text":"That just means that the function is a linear function."},{"Start":"01:29.120 ","End":"01:35.164","Text":"Second derivative is 0 for a continuous function means that it\u0027s a linear function,"},{"Start":"01:35.164 ","End":"01:37.310","Text":"so it doesn\u0027t have to be 0."},{"Start":"01:37.310 ","End":"01:38.795","Text":"It could, for example,"},{"Start":"01:38.795 ","End":"01:42.680","Text":"take f of x to be the constant function 1 just as an example,"},{"Start":"01:42.680 ","End":"01:44.930","Text":"then a second derivative would be 0."},{"Start":"01:44.930 ","End":"01:47.890","Text":"This would be 0,"},{"Start":"01:47.890 ","End":"01:52.200","Text":"but the function is not 0. We\u0027re done."}],"ID":28651},{"Watched":false,"Name":"Exercise 4","Duration":"3m 12s","ChapterTopicVideoID":27457,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"In this exercise, V is a space of"},{"Start":"00:03.510 ","End":"00:08.835","Text":"continuously differentiable real functions on the interval from minus 1,1."},{"Start":"00:08.835 ","End":"00:14.910","Text":"But they have to satisfy an additional condition that f(minus 1) is 0,"},{"Start":"00:14.910 ","End":"00:17.925","Text":"that seems unusual and we\u0027ll see why it\u0027s needed."},{"Start":"00:17.925 ","End":"00:23.895","Text":"We define an inner product on this space or if we define a product as follows,"},{"Start":"00:23.895 ","End":"00:27.540","Text":"we have to prove that this is indeed an inner product on"},{"Start":"00:27.540 ","End":"00:32.820","Text":"V. Here\u0027s a reminder of the axioms for inner product."},{"Start":"00:32.820 ","End":"00:36.570","Text":"I\u0027ll remind you here again what the product is."},{"Start":"00:36.570 ","End":"00:39.190","Text":"Now let\u0027s start with the axioms."},{"Start":"00:39.190 ","End":"00:42.905","Text":"Number 1, is called linearity in the first argument,"},{"Start":"00:42.905 ","End":"00:46.489","Text":"this is additivity and this is multiplication with a scalar."},{"Start":"00:46.489 ","End":"00:55.440","Text":"Now f plus g derivative is f derivative plus g derivative times h derivative."},{"Start":"00:55.440 ","End":"01:00.979","Text":"Then we use the distributive law and also that the integral is additive,"},{"Start":"01:00.979 ","End":"01:03.065","Text":"so we break it up like so."},{"Start":"01:03.065 ","End":"01:05.060","Text":"This is f,"},{"Start":"01:05.060 ","End":"01:06.590","Text":"h plus g,"},{"Start":"01:06.590 ","End":"01:11.010","Text":"h the scalar products, that\u0027s the additivity."},{"Start":"01:11.010 ","End":"01:13.110","Text":"Now multiplication with a scalar,"},{"Start":"01:13.110 ","End":"01:16.700","Text":"Alpha f derivative is Alpha times"},{"Start":"01:16.700 ","End":"01:20.710","Text":"f derivative and the Alpha comes in front of the integral sign."},{"Start":"01:20.710 ","End":"01:27.590","Text":"This is just the inner product of f and g. That\u0027s what we wanted, that\u0027s linearity."},{"Start":"01:27.590 ","End":"01:30.380","Text":"Next, we want to conjugate symmetry."},{"Start":"01:30.380 ","End":"01:35.104","Text":"This is so, now recall that if z is a real number,"},{"Start":"01:35.104 ","End":"01:39.950","Text":"then z-bar equals z as complex numbers."},{"Start":"01:39.950 ","End":"01:42.350","Text":"Let\u0027s take this side,"},{"Start":"01:42.350 ","End":"01:46.550","Text":"g inner product with f. Conjugate is just the same as"},{"Start":"01:46.550 ","End":"01:51.400","Text":"g product with f not to the integral of g(x), f(x)dx."},{"Start":"01:51.400 ","End":"01:53.520","Text":"We can switch the order of the product,"},{"Start":"01:53.520 ","End":"01:58.140","Text":"and then it gives us exactly f product g. Now the third axiom,"},{"Start":"01:58.140 ","End":"02:01.860","Text":"positive definiteness has 2 parts,"},{"Start":"02:01.860 ","End":"02:05.030","Text":"this u,u is bigger or equal to 0 and u,"},{"Start":"02:05.030 ","End":"02:07.505","Text":"u is 0 if and only if u is 0."},{"Start":"02:07.505 ","End":"02:10.610","Text":"Let\u0027s see, take f instead of u, f,"},{"Start":"02:10.610 ","End":"02:15.890","Text":"f is the integral minus 1,1 of (f\u0027)^2 dx,"},{"Start":"02:15.890 ","End":"02:18.155","Text":"and it\u0027s something squared,"},{"Start":"02:18.155 ","End":"02:21.260","Text":"so the integral is going to be bigger or equal to 0."},{"Start":"02:21.260 ","End":"02:25.650","Text":"Now, if f inner product with itself is 0,"},{"Start":"02:25.650 ","End":"02:28.170","Text":"then this integral is 0."},{"Start":"02:28.170 ","End":"02:31.065","Text":"This means that f\u0027 of x^2,"},{"Start":"02:31.065 ","End":"02:33.785","Text":"I don\u0027t know why I put absolute value here, but never mind."},{"Start":"02:33.785 ","End":"02:40.050","Text":"This is equal to 0 everywhere because it\u0027s non-negative and if the integral is 0,"},{"Start":"02:40.050 ","End":"02:42.055","Text":"then the function is 0 everywhere."},{"Start":"02:42.055 ","End":"02:45.275","Text":"Of course, this has to be continuous to deduce this."},{"Start":"02:45.275 ","End":"02:47.450","Text":"If something squared is 0 everywhere,"},{"Start":"02:47.450 ","End":"02:50.540","Text":"then itself is 0 everywhere on the interval,"},{"Start":"02:50.540 ","End":"02:54.605","Text":"that means that the function itself is a constant."},{"Start":"02:54.605 ","End":"02:59.915","Text":"Now, remember that condition that f(minus 1) is 0."},{"Start":"02:59.915 ","End":"03:05.185","Text":"That means that f(x) is always equal to f(minus 1) because it\u0027s a constant."},{"Start":"03:05.185 ","End":"03:07.440","Text":"F(x) is always 0,"},{"Start":"03:07.440 ","End":"03:09.810","Text":"which means that f is the 0 function."},{"Start":"03:09.810 ","End":"03:13.330","Text":"That\u0027s number 3, and we\u0027re done."}],"ID":28652},{"Watched":false,"Name":"Tutorial - Normed Space","Duration":"2m 20s","ChapterTopicVideoID":27446,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.650","Text":"In the previous clip,"},{"Start":"00:01.650 ","End":"00:03.750","Text":"we learned about inner product spaces."},{"Start":"00:03.750 ","End":"00:06.705","Text":"Now we\u0027ll learn what a normed space is."},{"Start":"00:06.705 ","End":"00:10.470","Text":"As before, it\u0027s going to be only over 2 fields,"},{"Start":"00:10.470 ","End":"00:14.130","Text":"C or R, complex or real."},{"Start":"00:14.130 ","End":"00:19.095","Text":"We also begin with a vector space over a field."},{"Start":"00:19.095 ","End":"00:25.470","Text":"A norm is a function from the vector space to the real."},{"Start":"00:25.470 ","End":"00:29.918","Text":"It assigns to each vector a real number,"},{"Start":"00:29.918 ","End":"00:34.260","Text":"and it\u0027s denoted as the norm of U, double bars."},{"Start":"00:34.260 ","End":"00:37.370","Text":"It has to satisfy the following axioms."},{"Start":"00:37.370 ","End":"00:41.405","Text":"First of all, the norm of anything is bigger or equal to 0,"},{"Start":"00:41.405 ","End":"00:46.459","Text":"and it\u0027s equal to 0 if and only if the vector is 0."},{"Start":"00:46.459 ","End":"00:51.035","Text":"Secondly, multiplication of the vector by a scalar"},{"Start":"00:51.035 ","End":"00:56.510","Text":"causes the norm to be multiplied by the absolute value of the scalar."},{"Start":"00:56.510 ","End":"01:01.100","Text":"This is for all scalars Alpha and for all vectors U."},{"Start":"01:01.100 ","End":"01:05.315","Text":"Also we have the triangle inequality that the norm of"},{"Start":"01:05.315 ","End":"01:09.433","Text":"u plus v is less than or equal to the norm of u plus the norm of v,"},{"Start":"01:09.433 ","End":"01:16.060","Text":"for all u and v. A vector space with a norm is called a normed space."},{"Start":"01:16.060 ","End":"01:20.029","Text":"It turns out that every inner product space is a normed space."},{"Start":"01:20.029 ","End":"01:22.180","Text":"There\u0027s something called an induced norm."},{"Start":"01:22.180 ","End":"01:23.752","Text":"If V is an inner product space,"},{"Start":"01:23.752 ","End":"01:27.470","Text":"then you can define a norm on v by the following."},{"Start":"01:27.470 ","End":"01:33.650","Text":"The norm of u will be the square root of u scalar product with itself."},{"Start":"01:33.650 ","End":"01:35.786","Text":"It turns out, and I won\u0027t prove it,"},{"Start":"01:35.786 ","End":"01:37.580","Text":"that this is a norm,"},{"Start":"01:37.580 ","End":"01:41.450","Text":"and it\u0027s actually the induced norm by the inner product."},{"Start":"01:41.450 ","End":"01:43.220","Text":"However, I\u0027ll remark that"},{"Start":"01:43.220 ","End":"01:47.795","Text":"not every normed space has its norm induced from an inner product."},{"Start":"01:47.795 ","End":"01:51.395","Text":"Normed spaces are more general than inner product spaces."},{"Start":"01:51.395 ","End":"01:52.715","Text":"They\u0027re more abundant."},{"Start":"01:52.715 ","End":"01:55.205","Text":"Inner product spaces are more specialized."},{"Start":"01:55.205 ","End":"01:59.038","Text":"It\u0027s a very useful inequality in normed spaces,"},{"Start":"01:59.038 ","End":"02:02.530","Text":"and it\u0027s called the Cauchy-Schwarz inequality."},{"Start":"02:02.530 ","End":"02:06.320","Text":"This is basically what it is that the absolute value of"},{"Start":"02:06.320 ","End":"02:11.215","Text":"the inner product is less than or equal to the product of the norms."},{"Start":"02:11.215 ","End":"02:13.040","Text":"Have it here for reference,"},{"Start":"02:13.040 ","End":"02:15.515","Text":"we\u0027ll be using it frequently."},{"Start":"02:15.515 ","End":"02:20.460","Text":"That concludes this introductory clip to normed spaces."}],"ID":28653},{"Watched":false,"Name":"Tutorial - Common Spaces","Duration":"4m 53s","ChapterTopicVideoID":27447,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.240","Text":"In this clip we\u0027ll meet some of the more popular norm spaces and inner product spaces."},{"Start":"00:06.240 ","End":"00:09.405","Text":"Perhaps popular isn\u0027t the right word,"},{"Start":"00:09.405 ","End":"00:12.405","Text":"maybe common or frequently used."},{"Start":"00:12.405 ","End":"00:15.600","Text":"Anyway, start with the first one,"},{"Start":"00:15.600 ","End":"00:21.525","Text":"L2, and then the PC stands for piecewise continuous."},{"Start":"00:21.525 ","End":"00:26.955","Text":"This is a space of all piecewise continuous functions on the interval a, b."},{"Start":"00:26.955 ","End":"00:29.475","Text":"This is not entirely accurate."},{"Start":"00:29.475 ","End":"00:32.640","Text":"To be precise, we say that two functions are the"},{"Start":"00:32.640 ","End":"00:37.313","Text":"same if they differ only on a finite number of points."},{"Start":"00:37.313 ","End":"00:41.375","Text":"But you can forget that for practical purposes."},{"Start":"00:41.375 ","End":"00:46.255","Text":"It\u0027s an inner product space with the inner product defined as follows,"},{"Start":"00:46.255 ","End":"00:50.735","Text":"f times g is the integral from a to b,"},{"Start":"00:50.735 ","End":"00:55.190","Text":"of f(x), g(x) conjugate dx."},{"Start":"00:55.190 ","End":"00:57.500","Text":"The functions could be real or complex."},{"Start":"00:57.500 ","End":"01:00.935","Text":"If it\u0027s real functions you don\u0027t need the conjugate here."},{"Start":"01:00.935 ","End":"01:03.980","Text":"Now it has a norm induced by the inner product."},{"Start":"01:03.980 ","End":"01:07.825","Text":"The norm of f is the square root of f.f,"},{"Start":"01:07.825 ","End":"01:10.710","Text":"and that\u0027s the square root of f,"},{"Start":"01:10.710 ","End":"01:15.260","Text":"f bar is the absolute value of f squared."},{"Start":"01:15.260 ","End":"01:20.960","Text":"Any complex number times its conjugate is the absolute value squared."},{"Start":"01:20.960 ","End":"01:25.570","Text":"The next one is L1_PC of ab."},{"Start":"01:25.570 ","End":"01:27.390","Text":"As far as the elements go,"},{"Start":"01:27.390 ","End":"01:28.430","Text":"it\u0027s the same as this."},{"Start":"01:28.430 ","End":"01:32.795","Text":"It\u0027s the space of all piecewise continuous functions on the interval a, b."},{"Start":"01:32.795 ","End":"01:37.145","Text":"The same remark is here, applies here also."},{"Start":"01:37.145 ","End":"01:40.020","Text":"But it doesn\u0027t have an inner product to normed space,"},{"Start":"01:40.020 ","End":"01:42.050","Text":"if we define the norm as follows,"},{"Start":"01:42.050 ","End":"01:46.625","Text":"norm of f is the integral from a to b absolute value of f dx."},{"Start":"01:46.625 ","End":"01:50.990","Text":"Turns out that it\u0027s not a very interesting or useful space,"},{"Start":"01:50.990 ","End":"01:57.280","Text":"but it\u0027s good as an example of a norm which is not induced by any inner product."},{"Start":"01:57.280 ","End":"02:03.200","Text":"Won\u0027t prove this, but it\u0027s known that this cannot be induced by an inner product."},{"Start":"02:03.200 ","End":"02:08.660","Text":"Next one, same as the first one except instead of the interval a,"},{"Start":"02:08.660 ","End":"02:11.760","Text":"b, we take the whole real line,"},{"Start":"02:11.760 ","End":"02:14.175","Text":"or the interval for minus infinity to infinity."},{"Start":"02:14.175 ","End":"02:18.120","Text":"But we can\u0027t take old piecewise continuous functions f,"},{"Start":"02:18.120 ","End":"02:22.084","Text":"we only take the ones which satisfy this inequality,"},{"Start":"02:22.084 ","End":"02:27.200","Text":"that the absolute value squared is integrable on minus infinity to infinity,"},{"Start":"02:27.200 ","End":"02:29.240","Text":"means it has a finite integral."},{"Start":"02:29.240 ","End":"02:32.569","Text":"A similar remark applies,"},{"Start":"02:32.569 ","End":"02:33.680","Text":"like here and here,"},{"Start":"02:33.680 ","End":"02:34.760","Text":"not quite the same,"},{"Start":"02:34.760 ","End":"02:36.560","Text":"but let\u0027s not get too technical,"},{"Start":"02:36.560 ","End":"02:38.660","Text":"for practical purposes this will do."},{"Start":"02:38.660 ","End":"02:44.230","Text":"On this space we can define an inner product and hence a norm as follows,"},{"Start":"02:44.230 ","End":"02:50.680","Text":"f times g is the integral from minus infinity to infinity of f times g conjugate."},{"Start":"02:50.680 ","End":"02:53.920","Text":"The norm is just the induced norm."},{"Start":"02:53.920 ","End":"02:58.430","Text":"It will be the square root of the integral of f,"},{"Start":"02:58.430 ","End":"03:02.165","Text":"f bar, which is the absolute value of f^2."},{"Start":"03:02.165 ","End":"03:06.310","Text":"Similar to the L2 space is an L1 space,"},{"Start":"03:06.310 ","End":"03:11.915","Text":"all piecewise continuous functions on the whole real line."},{"Start":"03:11.915 ","End":"03:16.025","Text":"I should have said here and here the functions could be complex-valued,"},{"Start":"03:16.025 ","End":"03:21.050","Text":"the defined for the real domain but the range can be complex."},{"Start":"03:21.050 ","End":"03:25.310","Text":"This one is piecewise continuous functions on the real line,"},{"Start":"03:25.310 ","End":"03:30.275","Text":"such that the following integral converges or is less than infinity."},{"Start":"03:30.275 ","End":"03:33.530","Text":"We define the norm doesn\u0027t have an inner product,"},{"Start":"03:33.530 ","End":"03:36.290","Text":"has a norm as follows."},{"Start":"03:36.290 ","End":"03:40.910","Text":"The next example, not functions but sequences."},{"Start":"03:40.910 ","End":"03:44.480","Text":"The space of all infinite sequences could be real or complex."},{"Start":"03:44.480 ","End":"03:47.450","Text":"Sequences x_n from 0 to infinity,"},{"Start":"03:47.450 ","End":"03:52.640","Text":"although strictly speaking it could be from 1 to infinity or from any place to infinity."},{"Start":"03:52.640 ","End":"03:56.315","Text":"Typically it\u0027s complex sequences,"},{"Start":"03:56.315 ","End":"03:59.405","Text":"and they have to satisfy that"},{"Start":"03:59.405 ","End":"04:05.530","Text":"the sum of the series of the absolute values converges, less than infinity."},{"Start":"04:05.530 ","End":"04:10.070","Text":"Well becomes a norm space if you define the following norm."},{"Start":"04:10.070 ","End":"04:15.245","Text":"The norm of a sequence x_n is the sum of the absolute values of x_n,"},{"Start":"04:15.245 ","End":"04:17.245","Text":"which will be finite."},{"Start":"04:17.245 ","End":"04:19.400","Text":"Then there\u0027s a similar one."},{"Start":"04:19.400 ","End":"04:23.722","Text":"The space of all infinite sequences satisfy a different condition,"},{"Start":"04:23.722 ","End":"04:28.880","Text":"this time the sum of the absolute value squared has to be finite."},{"Start":"04:28.880 ","End":"04:32.180","Text":"The inner product is as follows."},{"Start":"04:32.180 ","End":"04:34.830","Text":"The sequence x_n times a sequence y_n,"},{"Start":"04:34.830 ","End":"04:38.105","Text":"is the sum of x_n, y_n conjugate,"},{"Start":"04:38.105 ","End":"04:42.131","Text":"and the induced norm is the following,"},{"Start":"04:42.131 ","End":"04:46.670","Text":"square root of the sum of x_n and absolute value squared."},{"Start":"04:46.670 ","End":"04:53.760","Text":"That will do for the collection of well-known spaces. We are done."}],"ID":28654},{"Watched":false,"Name":"Exercise 5","Duration":"2m 18s","ChapterTopicVideoID":27448,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.760","Text":"This exercise serves mostly as an example of a normed space."},{"Start":"00:05.760 ","End":"00:11.040","Text":"Let V be the space of continuous functions on the interval a, b,"},{"Start":"00:11.040 ","End":"00:12.690","Text":"and we have to prove that"},{"Start":"00:12.690 ","End":"00:17.655","Text":"the expression norm of f is the integral from a to b of absolute value of f(x),"},{"Start":"00:17.655 ","End":"00:23.295","Text":"dx defines a norm on V. This is a remark."},{"Start":"00:23.295 ","End":"00:27.810","Text":"It\u0027s often denoted as norm with a subscript 1."},{"Start":"00:27.810 ","End":"00:30.770","Text":"Anyway, here are the axioms for"},{"Start":"00:30.770 ","End":"00:35.390","Text":"a normed space and we\u0027ll show that these axioms hold in our case."},{"Start":"00:35.390 ","End":"00:40.100","Text":"The first part, we have to show that it\u0027s non-negative."},{"Start":"00:40.100 ","End":"00:43.790","Text":"The norm of f is the integral of the absolute value of something,"},{"Start":"00:43.790 ","End":"00:45.800","Text":"and so of course is non-negative."},{"Start":"00:45.800 ","End":"00:49.820","Text":"Next, we have to show that the norm is 0 if and only if the vector,"},{"Start":"00:49.820 ","End":"00:51.875","Text":"in this case the function, is 0."},{"Start":"00:51.875 ","End":"00:54.805","Text":"Norm of f is 0 if and only if the integral from a"},{"Start":"00:54.805 ","End":"00:58.280","Text":"to b of the absolute value of f(x) is 0."},{"Start":"00:58.280 ","End":"01:01.085","Text":"Now, here it\u0027s important that f is continuous,"},{"Start":"01:01.085 ","End":"01:04.100","Text":"and so is the absolute value and the integral of"},{"Start":"01:04.100 ","End":"01:09.905","Text":"a non-negative function if it\u0027s 0 and the function has to be 0."},{"Start":"01:09.905 ","End":"01:13.205","Text":"In brief, if it isn\u0027t always 0,"},{"Start":"01:13.205 ","End":"01:17.180","Text":"and at some point it\u0027s strictly positive the absolute value."},{"Start":"01:17.180 ","End":"01:21.110","Text":"There\u0027s an interval where the absolute value is positive,"},{"Start":"01:21.110 ","End":"01:23.570","Text":"so the integral is positive on that interval."},{"Start":"01:23.570 ","End":"01:25.220","Text":"If we extend it to all of a, b,"},{"Start":"01:25.220 ","End":"01:29.524","Text":"it can only get bigger so it\u0027s bigger than 0."},{"Start":"01:29.524 ","End":"01:33.485","Text":"The other one, multiplication by a scalar."},{"Start":"01:33.485 ","End":"01:39.125","Text":"The norm of Alpha f is the integral of Alpha times f(x) absolute value."},{"Start":"01:39.125 ","End":"01:44.435","Text":"We can take Alpha out in absolute value and bring it in front of the integral."},{"Start":"01:44.435 ","End":"01:49.849","Text":"This is exactly absolute value of Alpha times the norm of f. Next,"},{"Start":"01:49.849 ","End":"01:52.315","Text":"we have the triangle inequality."},{"Start":"01:52.315 ","End":"01:54.410","Text":"Let\u0027s see about that one."},{"Start":"01:54.410 ","End":"01:58.280","Text":"The norm of f plus g is the integral of absolute value of"},{"Start":"01:58.280 ","End":"02:02.480","Text":"f plus g. Then by the regular triangle inequality,"},{"Start":"02:02.480 ","End":"02:07.250","Text":"this is less than or equal to absolute value of f plus absolute value of g here."},{"Start":"02:07.250 ","End":"02:11.140","Text":"Then the integral of the sum is the sum of the integrals."},{"Start":"02:11.140 ","End":"02:18.400","Text":"This is the norm of f plus the norm of g. That concludes this exercise."}],"ID":28655},{"Watched":false,"Name":"Exercise 6","Duration":"2m 21s","ChapterTopicVideoID":27449,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.920","Text":"In this exercise, we present another example of a norm space."},{"Start":"00:04.920 ","End":"00:09.480","Text":"Again, the space of continuous functions on the interval a, b."},{"Start":"00:09.480 ","End":"00:12.735","Text":"This time the norm is the maximum."},{"Start":"00:12.735 ","End":"00:13.830","Text":"It\u0027s a supremum,"},{"Start":"00:13.830 ","End":"00:17.010","Text":"but it\u0027s the same as the maximum because the function is continuous,"},{"Start":"00:17.010 ","End":"00:18.780","Text":"maximum on the interval."},{"Start":"00:18.780 ","End":"00:21.510","Text":"We have to show that this defines a norm really."},{"Start":"00:21.510 ","End":"00:25.190","Text":"By the way, this is often denoted as a norm"},{"Start":"00:25.190 ","End":"00:29.180","Text":"with a subscript infinity, this particular norm."},{"Start":"00:29.180 ","End":"00:34.235","Text":"Here are the axioms and let\u0027s see if we can verify the axioms."},{"Start":"00:34.235 ","End":"00:36.230","Text":"First of all, non-negative."},{"Start":"00:36.230 ","End":"00:39.590","Text":"Well, that\u0027s clear because the maximum of the absolute value,"},{"Start":"00:39.590 ","End":"00:41.270","Text":"it\u0027s got to be non-negative."},{"Start":"00:41.270 ","End":"00:44.630","Text":"Then suppose norm of f is 0,"},{"Start":"00:44.630 ","End":"00:47.105","Text":"that means that the maximum is 0."},{"Start":"00:47.105 ","End":"00:51.530","Text":"Maximum of the absolute value of the function is 0 and it\u0027s non-negative,"},{"Start":"00:51.530 ","End":"00:54.560","Text":"then it has to be 0 everywhere the absolute value,"},{"Start":"00:54.560 ","End":"00:57.080","Text":"and so the function itself also is 0,"},{"Start":"00:57.080 ","End":"00:59.390","Text":"and it works the other way round also."},{"Start":"00:59.390 ","End":"01:03.845","Text":"The function is 0, then the maximum of the absolute value is 0."},{"Start":"01:03.845 ","End":"01:07.690","Text":"Next is the scalar multiple."},{"Start":"01:07.690 ","End":"01:14.015","Text":"Norm of Alpha f is the maximum of Alpha times f(x) absolute value."},{"Start":"01:14.015 ","End":"01:17.690","Text":"We can take the absolute value of Alpha,"},{"Start":"01:17.690 ","End":"01:21.050","Text":"split it from the absolute value of f(x)."},{"Start":"01:21.050 ","End":"01:23.000","Text":"Then because this is non-negative,"},{"Start":"01:23.000 ","End":"01:25.175","Text":"we can take it in front of the maximum."},{"Start":"01:25.175 ","End":"01:27.980","Text":"If it was negative, we\u0027d have to make this a minimum."},{"Start":"01:27.980 ","End":"01:33.605","Text":"We have this, but this is just the norm of f. We\u0027ve got what we wanted here."},{"Start":"01:33.605 ","End":"01:36.505","Text":"What\u0027s left is the triangle inequality."},{"Start":"01:36.505 ","End":"01:40.190","Text":"The norm of f plus g is the maximum of absolute value of f plus"},{"Start":"01:40.190 ","End":"01:44.890","Text":"g. Absolute value of the sum is the sum of the absolute values."},{"Start":"01:44.890 ","End":"01:50.150","Text":"This is going to be less than or equal to absolute value of f plus absolute value of"},{"Start":"01:50.150 ","End":"01:55.730","Text":"g. Now I claim that we can take the maximum of each of these separately,"},{"Start":"01:55.730 ","End":"02:01.340","Text":"because absolute value of f(x) is less than or equal to the maximum for any x,"},{"Start":"02:01.340 ","End":"02:04.340","Text":"and the absolute value of g is less than or equal to maximum."},{"Start":"02:04.340 ","End":"02:09.025","Text":"For any x, this plus this is less than or equal to the maximum plus the maximum."},{"Start":"02:09.025 ","End":"02:11.630","Text":"If for any x this is true when we take the maximum,"},{"Start":"02:11.630 ","End":"02:13.325","Text":"it\u0027s still going to be true."},{"Start":"02:13.325 ","End":"02:16.700","Text":"The sum is less than or equal to the sum of the maximum."},{"Start":"02:16.700 ","End":"02:22.440","Text":"This is just the norm of f plus the norm of g. That concludes this exercise."}],"ID":28656},{"Watched":false,"Name":"Exercise 7 - Part a","Duration":"1m 51s","ChapterTopicVideoID":27451,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:03.780","Text":"In this exercise, V is an inner product space"},{"Start":"00:03.780 ","End":"00:07.365","Text":"and also in a normed space with the induced norm."},{"Start":"00:07.365 ","End":"00:08.640","Text":"Let u, v,"},{"Start":"00:08.640 ","End":"00:14.445","Text":"w be vectors in V. We have to prove for things and we\u0027ll do each one in a separate clip."},{"Start":"00:14.445 ","End":"00:16.050","Text":"Start with a;"},{"Start":"00:16.050 ","End":"00:19.260","Text":"u times v plus w is u,"},{"Start":"00:19.260 ","End":"00:22.200","Text":"v plus u, w. However,"},{"Start":"00:22.200 ","End":"00:24.915","Text":"u times Alpha w,"},{"Start":"00:24.915 ","End":"00:27.030","Text":"I should\u0027ve said Alpha is a scalar,"},{"Start":"00:27.030 ","End":"00:30.089","Text":"Alpha is in the field real or complex,"},{"Start":"00:30.089 ","End":"00:35.150","Text":"is Alpha conjugate times u,"},{"Start":"00:35.150 ","End":"00:40.410","Text":"v. This is called semi-linearity in the second argument."},{"Start":"00:40.410 ","End":"00:41.750","Text":"In the first argument,"},{"Start":"00:41.750 ","End":"00:43.670","Text":"Alpha comes out as it is,"},{"Start":"00:43.670 ","End":"00:47.910","Text":"but in the second argument it comes out with a conjugate."},{"Start":"00:47.910 ","End":"00:50.595","Text":"Let\u0027s solve part a,"},{"Start":"00:50.595 ","End":"00:57.065","Text":"u times v plus w. First of all we can flip them around and put a conjugate on the top."},{"Start":"00:57.065 ","End":"01:02.690","Text":"Then we can use the linearity of the first argument to say this is v,"},{"Start":"01:02.690 ","End":"01:03.805","Text":"u plus w, u."},{"Start":"01:03.805 ","End":"01:05.430","Text":"Again, under a conjugate."},{"Start":"01:05.430 ","End":"01:08.640","Text":"Conjugate of a sum is sum of the conjugates and then"},{"Start":"01:08.640 ","End":"01:11.980","Text":"we can flip each one around again because of the conjugate and get u,"},{"Start":"01:11.980 ","End":"01:16.445","Text":"v plus u, w. That\u0027s this part of the linearity."},{"Start":"01:16.445 ","End":"01:20.630","Text":"Now the semi linearity with the scalar in the second argument,"},{"Start":"01:20.630 ","End":"01:23.560","Text":"u come Alpha v. Again,"},{"Start":"01:23.560 ","End":"01:26.975","Text":"flip the order and put a roof on the top."},{"Start":"01:26.975 ","End":"01:30.590","Text":"This is equal to Alpha comes out as it is."},{"Start":"01:30.590 ","End":"01:33.260","Text":"It\u0027s in the first argument. Then we can say"},{"Start":"01:33.260 ","End":"01:36.515","Text":"the conjugate of a product is the product of the conjugates."},{"Start":"01:36.515 ","End":"01:38.653","Text":"It\u0027s Alpha conjugate times v,"},{"Start":"01:38.653 ","End":"01:44.420","Text":"u conjugate and then we can flip the order here and get u comma v,"},{"Start":"01:44.420 ","End":"01:47.720","Text":"but Alpha stays conjugate but this is what we wanted."},{"Start":"01:47.720 ","End":"01:52.440","Text":"That\u0027s part a, and that\u0027s this clip."}],"ID":28657},{"Watched":false,"Name":"Exercise 7 - Part b","Duration":"1m 39s","ChapterTopicVideoID":27452,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"We just did part a,"},{"Start":"00:01.890 ","End":"00:04.620","Text":"we\u0027re now going to do part b of this exercise."},{"Start":"00:04.620 ","End":"00:07.845","Text":"We have to prove the following equality."},{"Start":"00:07.845 ","End":"00:12.750","Text":"Again, we\u0027re an inner product space and u and v are vectors in it."},{"Start":"00:12.750 ","End":"00:15.435","Text":"We have to prove this identity."},{"Start":"00:15.435 ","End":"00:20.273","Text":"Let\u0027s start with the right-hand side and see if we can get to the left-hand side,"},{"Start":"00:20.273 ","End":"00:25.580","Text":"1/4 of normal u plus v^2 minus norm of u minus v^2 is 1/4."},{"Start":"00:25.580 ","End":"00:29.000","Text":"The norm of something squared is the dot-product"},{"Start":"00:29.000 ","End":"00:32.630","Text":"of the thing with itself so we have this dot-product,"},{"Start":"00:32.630 ","End":"00:35.540","Text":"scalar product in a product, same thing."},{"Start":"00:35.540 ","End":"00:40.010","Text":"Then we get just 4 things by the linearity,"},{"Start":"00:40.010 ","End":"00:41.930","Text":"this with this, this with this, anyway,"},{"Start":"00:41.930 ","End":"00:44.940","Text":"we get these 4, and here we get"},{"Start":"00:44.940 ","End":"00:49.940","Text":"minus these 4 and I\u0027ve taken care of the minus, minus is plus."},{"Start":"00:49.940 ","End":"00:52.250","Text":"Now stuff cancels, u,"},{"Start":"00:52.250 ","End":"00:55.765","Text":"u minus u, u cancels."},{"Start":"00:55.765 ","End":"01:02.450","Text":"Then v, v cancels with minus v, v. Well,"},{"Start":"01:02.450 ","End":"01:04.940","Text":"we\u0027re left with is this and another one of these,"},{"Start":"01:04.940 ","End":"01:08.015","Text":"and then this and another one of those like so,"},{"Start":"01:08.015 ","End":"01:10.835","Text":"then we can combine the same ones."},{"Start":"01:10.835 ","End":"01:12.800","Text":"This and this is 2 of these."},{"Start":"01:12.800 ","End":"01:13.910","Text":"This, and this is 2 of these."},{"Start":"01:13.910 ","End":"01:18.335","Text":"We can take the 2 out and it cancels with the 4 and we get 1.5 of this plus this."},{"Start":"01:18.335 ","End":"01:20.590","Text":"Next we can write this plus this over 2,"},{"Start":"01:20.590 ","End":"01:22.300","Text":"but v, u is u,"},{"Start":"01:22.300 ","End":"01:26.000","Text":"v with the conjugate so we have this now that the result from"},{"Start":"01:26.000 ","End":"01:33.035","Text":"complex numbers that z plus z conjugate over 2 is the real part of z."},{"Start":"01:33.035 ","End":"01:35.735","Text":"Here we just get the real part of u,"},{"Start":"01:35.735 ","End":"01:40.200","Text":"v, and that\u0027s what we wanted and so that\u0027s part b."}],"ID":28658},{"Watched":false,"Name":"Exercise 7 - Part c","Duration":"2m 14s","ChapterTopicVideoID":27453,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"Now we\u0027ve done part a and b of this exercise,"},{"Start":"00:03.480 ","End":"00:06.675","Text":"we\u0027re up to part c. Let\u0027s start with it."},{"Start":"00:06.675 ","End":"00:09.460","Text":"We have to prove this identity."},{"Start":"00:09.460 ","End":"00:11.300","Text":"It\u0027s similar to part b,"},{"Start":"00:11.300 ","End":"00:13.220","Text":"so we can go a bit quicker."},{"Start":"00:13.220 ","End":"00:17.670","Text":"We have 1/4 of this then we can expand because norm squared"},{"Start":"00:17.670 ","End":"00:22.200","Text":"is the scalar product of the thing with itself."},{"Start":"00:22.200 ","End":"00:27.780","Text":"Then we can use the distributive law here and we"},{"Start":"00:27.780 ","End":"00:33.746","Text":"get for this 4 pieces and for here another 4 pieces,"},{"Start":"00:33.746 ","End":"00:37.535","Text":"and take care of the minuses and pluses properly."},{"Start":"00:37.535 ","End":"00:40.520","Text":"Some things cancel; u,"},{"Start":"00:40.520 ","End":"00:43.955","Text":"u cancels with the minus u, u,"},{"Start":"00:43.955 ","End":"00:51.425","Text":"and also iv cancels with minus iv,"},{"Start":"00:51.425 ","End":"00:55.820","Text":"iv and we get this."},{"Start":"00:55.820 ","End":"00:58.550","Text":"But notice that this is repeated here and here,"},{"Start":"00:58.550 ","End":"01:00.425","Text":"and this is repeated here and here."},{"Start":"01:00.425 ","End":"01:03.680","Text":"We can take twice of this and twice of this,"},{"Start":"01:03.680 ","End":"01:06.595","Text":"bring a 2 out and the 2/4 is 1/2,"},{"Start":"01:06.595 ","End":"01:08.660","Text":"so we have this."},{"Start":"01:09.060 ","End":"01:17.855","Text":"Then we can write this as the opposite order and a conjugate on top."},{"Start":"01:17.855 ","End":"01:27.610","Text":"We can take a scalar from the first argument outside,"},{"Start":"01:27.610 ","End":"01:30.023","Text":"take the i outside here and here,"},{"Start":"01:30.023 ","End":"01:32.940","Text":"but it\u0027s still under the conjugate roof,"},{"Start":"01:32.940 ","End":"01:39.140","Text":"and then this will equal i conjugate is minus i."},{"Start":"01:39.140 ","End":"01:41.965","Text":"Do several things to get from here to here."},{"Start":"01:41.965 ","End":"01:43.300","Text":"First of all, v,"},{"Start":"01:43.300 ","End":"01:44.860","Text":"u conjugate is u,"},{"Start":"01:44.860 ","End":"01:47.200","Text":"v and v,"},{"Start":"01:47.200 ","End":"01:50.495","Text":"u without the conjugate is u, v conjugate."},{"Start":"01:50.495 ","End":"01:54.750","Text":"Next, we can multiply this by i and we get 1,"},{"Start":"01:54.750 ","End":"01:56.550","Text":"this by i, we get minus 1,"},{"Start":"01:56.550 ","End":"02:00.140","Text":"and we put another i in the denominator to even it out."},{"Start":"02:00.140 ","End":"02:03.680","Text":"Then we use the formula for complex numbers,"},{"Start":"02:03.680 ","End":"02:06.507","Text":"this one, and apply it to our case,"},{"Start":"02:06.507 ","End":"02:09.920","Text":"and we\u0027ve got the imaginary part of u,"},{"Start":"02:09.920 ","End":"02:14.910","Text":"v. That concludes part c."}],"ID":28659},{"Watched":false,"Name":"Exercise 7 - Part d","Duration":"1m 13s","ChapterTopicVideoID":27454,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.230 ","End":"00:03.255","Text":"Now we come to part d,"},{"Start":"00:03.255 ","End":"00:09.825","Text":"which is the parallelogram equality and the little diagram here."},{"Start":"00:09.825 ","End":"00:12.900","Text":"It\u0027s a generalized Pythagoras."},{"Start":"00:12.900 ","End":"00:14.939","Text":"If we have a parallelogram,"},{"Start":"00:14.939 ","End":"00:18.570","Text":"then the sum of the squares on the 4 sides"},{"Start":"00:18.570 ","End":"00:22.590","Text":"equals the sums of the squares on the 2 diagonals."},{"Start":"00:22.590 ","End":"00:24.960","Text":"That\u0027s really what it says."},{"Start":"00:24.960 ","End":"00:27.910","Text":"Let\u0027s prove that."},{"Start":"00:28.100 ","End":"00:31.320","Text":"As usual, when we see the norm squared,"},{"Start":"00:31.320 ","End":"00:35.655","Text":"we can write it as the vector.product with itself,"},{"Start":"00:35.655 ","End":"00:37.995","Text":"scalar product in the product."},{"Start":"00:37.995 ","End":"00:40.260","Text":"We get this expression."},{"Start":"00:40.260 ","End":"00:46.625","Text":"Then we expand and we get for each of these 4 terms and another 4."},{"Start":"00:46.625 ","End":"00:48.950","Text":"Then some things cancel."},{"Start":"00:48.950 ","End":"00:53.565","Text":"U, v cancels with minus u, v,"},{"Start":"00:53.565 ","End":"00:58.460","Text":"and v, u cancels with minus v, u."},{"Start":"00:58.460 ","End":"01:03.260","Text":"We\u0027re just left with twice u, u plus v,"},{"Start":"01:03.260 ","End":"01:08.925","Text":"v. But this is exactly norm of u^2 and this is norm of v^2."},{"Start":"01:08.925 ","End":"01:13.800","Text":"That concludes Part D. We\u0027re done."}],"ID":28660},{"Watched":false,"Name":"Exercise 8","Duration":"3m 22s","ChapterTopicVideoID":27458,"CourseChapterTopicPlaylistID":274166,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.660","Text":"In this exercise, V is the space of continuous complex functions on the interval a, b."},{"Start":"00:06.660 ","End":"00:10.605","Text":"We have to prove that the expression norm of f,"},{"Start":"00:10.605 ","End":"00:15.210","Text":"or we have to show that it defines a norm equals the integral from"},{"Start":"00:15.210 ","End":"00:20.110","Text":"a to b of absolute value of f(x)dx plus the maximum on a,"},{"Start":"00:20.110 ","End":"00:25.845","Text":"b of absolute value of f(x) defines a norm on V. Now,"},{"Start":"00:25.845 ","End":"00:30.450","Text":"we proved earlier that this part here is a norm,"},{"Start":"00:30.450 ","End":"00:34.170","Text":"and we called it f_1, a norm 1."},{"Start":"00:34.170 ","End":"00:38.555","Text":"We also defined this as norm infinity of f,"},{"Start":"00:38.555 ","End":"00:43.550","Text":"which is the maximum or supremum on the interval a, b."},{"Start":"00:43.550 ","End":"00:46.045","Text":"We showed that each of these is the norm."},{"Start":"00:46.045 ","End":"00:49.900","Text":"Our norm f is the sum of these 2."},{"Start":"00:49.900 ","End":"00:52.250","Text":"Now, it turns out that in general,"},{"Start":"00:52.250 ","End":"00:54.605","Text":"the sum of 2 norms is a norm."},{"Start":"00:54.605 ","End":"00:58.400","Text":"But we\u0027ll prove it for our specific case,"},{"Start":"00:58.400 ","End":"01:01.190","Text":"and you\u0027ll see that it\u0027s totally general that it would"},{"Start":"01:01.190 ","End":"01:04.325","Text":"work instead of f_1 plus f infinity."},{"Start":"01:04.325 ","End":"01:06.950","Text":"Here, you could put any a and b here,"},{"Start":"01:06.950 ","End":"01:09.410","Text":"or 1 and 2, or whatever."},{"Start":"01:09.410 ","End":"01:13.720","Text":"These are the axioms of a norm as a reminder."},{"Start":"01:13.720 ","End":"01:16.365","Text":"We\u0027ll go with the first one."},{"Start":"01:16.365 ","End":"01:20.210","Text":"Norm of f is norm 1 plus norm infinity of"},{"Start":"01:20.210 ","End":"01:24.250","Text":"f. Each of these is non-negative because each of these is a norm."},{"Start":"01:24.250 ","End":"01:27.140","Text":"If we add 2 non-negative numbers,"},{"Start":"01:27.140 ","End":"01:28.535","Text":"we get a non-negative."},{"Start":"01:28.535 ","End":"01:30.095","Text":"That\u0027s this part."},{"Start":"01:30.095 ","End":"01:32.165","Text":"Now, this part,"},{"Start":"01:32.165 ","End":"01:34.460","Text":"there\u0027s 2 directions for the arrow,"},{"Start":"01:34.460 ","End":"01:36.610","Text":"we\u0027ll do each one of them."},{"Start":"01:36.610 ","End":"01:40.935","Text":"Start from the right if a function f is 0,"},{"Start":"01:40.935 ","End":"01:43.320","Text":"then this norm 1 is a norm,"},{"Start":"01:43.320 ","End":"01:44.540","Text":"so it\u0027s 0,"},{"Start":"01:44.540 ","End":"01:48.490","Text":"and norm infinity of f is 0, both of them."},{"Start":"01:48.490 ","End":"01:52.595","Text":"The sum of them is 0 plus 0, which is 0."},{"Start":"01:52.595 ","End":"01:56.465","Text":"But this is exactly the norm of f is 0."},{"Start":"01:56.465 ","End":"01:57.980","Text":"In the other direction,"},{"Start":"01:57.980 ","End":"01:59.270","Text":"if we start here,"},{"Start":"01:59.270 ","End":"02:00.860","Text":"norm of f is 0,"},{"Start":"02:00.860 ","End":"02:03.035","Text":"so the sum of these 2 is 0."},{"Start":"02:03.035 ","End":"02:05.900","Text":"But these are non-negative real numbers,"},{"Start":"02:05.900 ","End":"02:07.685","Text":"and if the sum is 0,"},{"Start":"02:07.685 ","End":"02:09.830","Text":"they have to each be 0."},{"Start":"02:09.830 ","End":"02:11.670","Text":"But I could just take one of them,"},{"Start":"02:11.670 ","End":"02:13.200","Text":"I just need one of them to be 0,"},{"Start":"02:13.200 ","End":"02:14.550","Text":"say the first one."},{"Start":"02:14.550 ","End":"02:17.750","Text":"This one is 0. Because this is a norm,"},{"Start":"02:17.750 ","End":"02:19.520","Text":"it follows that f is 0."},{"Start":"02:19.520 ","End":"02:22.345","Text":"It could have equally well gone with the other one."},{"Start":"02:22.345 ","End":"02:25.850","Text":"Now, the next axiom multiplication with a scalar,"},{"Start":"02:25.850 ","End":"02:32.465","Text":"we have to show that this norm is the absolute value of Alpha times the norm of f,"},{"Start":"02:32.465 ","End":"02:35.050","Text":"which equal to the sum of these 2."},{"Start":"02:35.050 ","End":"02:40.085","Text":"Here we can take absolute value of Alpha outside the norm, and here also."},{"Start":"02:40.085 ","End":"02:42.635","Text":"Then by distributivity, we can"},{"Start":"02:42.635 ","End":"02:46.425","Text":"take absolute value of Alpha times this norm plus this norm."},{"Start":"02:46.425 ","End":"02:51.285","Text":"But this plus this is just the norm of f. We\u0027ve got what we wanted,"},{"Start":"02:51.285 ","End":"02:52.955","Text":"and that\u0027s this one."},{"Start":"02:52.955 ","End":"02:57.210","Text":"Now, last one, the triangle inequality."},{"Start":"02:57.210 ","End":"03:01.850","Text":"Norm of f plus g is norm 1 plus norm infinity."},{"Start":"03:01.850 ","End":"03:05.165","Text":"But the triangle inequality for this norm,"},{"Start":"03:05.165 ","End":"03:07.040","Text":"this is less than this plus this,"},{"Start":"03:07.040 ","End":"03:10.295","Text":"and this is less than or equal to this plus this, so we get this."},{"Start":"03:10.295 ","End":"03:13.955","Text":"Now if we put these 2 first and these 2 last,"},{"Start":"03:13.955 ","End":"03:20.484","Text":"we get norm of f plus norm of g. That\u0027s the last axiom,"},{"Start":"03:20.484 ","End":"03:22.540","Text":"and so we\u0027re done."}],"ID":28661}],"Thumbnail":null,"ID":274166},{"Name":"Convergence of Functions in Normed Spaces","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Pointwise Convergence","Duration":"4m 17s","ChapterTopicVideoID":27509,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.510","Text":"New topic, pointwise convergence of"},{"Start":"00:03.510 ","End":"00:08.489","Text":"functions is actually a more general concept of convergence of functions."},{"Start":"00:08.489 ","End":"00:12.255","Text":"This one is a specific kind called pointwise."},{"Start":"00:12.255 ","End":"00:16.525","Text":"You\u0027ve probably only encountered convergence of numbers,"},{"Start":"00:16.525 ","End":"00:20.295","Text":"and you may not have encountered convergence of functions."},{"Start":"00:20.295 ","End":"00:23.985","Text":"Just to recap, convergence of a sequence of numbers,"},{"Start":"00:23.985 ","End":"00:27.045","Text":"take for example the following sequence,"},{"Start":"00:27.045 ","End":"00:29.940","Text":"which we could describe as one over n,"},{"Start":"00:29.940 ","End":"00:32.454","Text":"where n=1, 2, 3, 4, etc."},{"Start":"00:32.454 ","End":"00:34.785","Text":"Natural numbers."},{"Start":"00:34.785 ","End":"00:37.380","Text":"We can talk about the limit,"},{"Start":"00:37.380 ","End":"00:39.140","Text":"in this case, obviously,"},{"Start":"00:39.140 ","End":"00:41.255","Text":"the limit is 0."},{"Start":"00:41.255 ","End":"00:44.780","Text":"There\u0027s also a convergence of a sequence of functions."},{"Start":"00:44.780 ","End":"00:46.580","Text":"Let\u0027s take, for example,"},{"Start":"00:46.580 ","End":"00:48.770","Text":"the following sequence x,"},{"Start":"00:48.770 ","End":"00:50.675","Text":"x2, x3, x4."},{"Start":"00:50.675 ","End":"00:54.560","Text":"Well, these are not just abstract polynomials, they\u0027re functions,"},{"Start":"00:54.560 ","End":"01:01.680","Text":"so we can write this more accurately as a sequence f_n(x)= x^n,"},{"Start":"01:01.680 ","End":"01:04.430","Text":"and x belongs to some domain I."},{"Start":"01:04.430 ","End":"01:08.150","Text":"For example, the interval from 0-1,"},{"Start":"01:08.150 ","End":"01:10.250","Text":"or the whole real line."},{"Start":"01:10.250 ","End":"01:16.490","Text":"Then we might ask, what is the limit as n goes to infinity of f_n(x)?"},{"Start":"01:16.490 ","End":"01:19.880","Text":"When we ask it this way and we choose a particular x,"},{"Start":"01:19.880 ","End":"01:23.030","Text":"we get what is called pointwise convergence."},{"Start":"01:23.030 ","End":"01:28.025","Text":"In general, we have a sequence f_n of functions,"},{"Start":"01:28.025 ","End":"01:31.760","Text":"and here we\u0027ll only be talking about real or complex functions defined on"},{"Start":"01:31.760 ","End":"01:37.490","Text":"some domain I to be a subset of the reals or the complex numbers."},{"Start":"01:37.490 ","End":"01:42.680","Text":"We said that the sequence f_n converges pointwise to f,"},{"Start":"01:42.680 ","End":"01:46.670","Text":"another function also defined on the same domain."},{"Start":"01:46.670 ","End":"01:50.135","Text":"If for all x in the domain,"},{"Start":"01:50.135 ","End":"01:58.765","Text":"the sequence of numbers f_n(x) converges as n goes to infinity to the number f(x)."},{"Start":"01:58.765 ","End":"02:04.430","Text":"We can write then that the limit as n goes to infinity of f_n is"},{"Start":"02:04.430 ","End":"02:10.130","Text":"f. This is the limit of functions equals a function and we write the word pointwise."},{"Start":"02:10.130 ","End":"02:12.230","Text":"I don\u0027t know there\u0027s a symbol in mathematical point."},{"Start":"02:12.230 ","End":"02:16.290","Text":"Well, you just write the word pointwise or you write f_n,"},{"Start":"02:16.290 ","End":"02:18.045","Text":"and with an arrow like this,"},{"Start":"02:18.045 ","End":"02:20.205","Text":"it goes to f pointwise."},{"Start":"02:20.205 ","End":"02:24.950","Text":"The function f is the pointwise limit of the sequence f_n."},{"Start":"02:24.950 ","End":"02:28.255","Text":"Now let\u0027s do a little example exercise."},{"Start":"02:28.255 ","End":"02:34.880","Text":"We have to check the pointwise convergence of the sequence f_n(x)= x^n,"},{"Start":"02:34.880 ","End":"02:36.905","Text":"the same one from the example."},{"Start":"02:36.905 ","End":"02:40.480","Text":"Here, I will be the closed interval 0,1."},{"Start":"02:40.480 ","End":"02:42.110","Text":"And if it converges,"},{"Start":"02:42.110 ","End":"02:44.915","Text":"we have to find the pointwise limit."},{"Start":"02:44.915 ","End":"02:47.825","Text":"If it doesn\u0027t converge, then no."},{"Start":"02:47.825 ","End":"02:52.670","Text":"Solution. Let\u0027s separate cases, take x=1 separately."},{"Start":"02:52.670 ","End":"02:53.720","Text":"If x= 1,"},{"Start":"02:53.720 ","End":"02:55.440","Text":"then x^n= 1,"},{"Start":"02:55.440 ","End":"02:56.685","Text":"whatever n is,"},{"Start":"02:56.685 ","End":"03:00.335","Text":"so the limit as n goes to infinity of x^n is 1,"},{"Start":"03:00.335 ","End":"03:01.850","Text":"and the limit is what we called f,"},{"Start":"03:01.850 ","End":"03:04.000","Text":"so f(1) is 1."},{"Start":"03:04.000 ","End":"03:09.260","Text":"As for the rest of the numbers that\u0027s bigger or equal to 0 and less than one,"},{"Start":"03:09.260 ","End":"03:11.680","Text":"x^n goes to 0."},{"Start":"03:11.680 ","End":"03:14.270","Text":"If x is less than 1, then x^n gets smaller and"},{"Start":"03:14.270 ","End":"03:17.780","Text":"smaller and smaller as n goes to infinity and it goes to 0."},{"Start":"03:17.780 ","End":"03:20.060","Text":"If x is 0, it\u0027s always 0."},{"Start":"03:20.060 ","End":"03:23.980","Text":"Any case, f(x) is 0, that\u0027s the limit."},{"Start":"03:23.980 ","End":"03:29.354","Text":"We can write this limit as piecewise, f(x) is 0."},{"Start":"03:29.354 ","End":"03:36.690","Text":"If x is between 0 and 1 excluding 1 and 1, if x=1."},{"Start":"03:36.690 ","End":"03:41.915","Text":"You can write that the limit as n goes to infinity of f_n, is f pointwise."},{"Start":"03:41.915 ","End":"03:44.885","Text":"Here\u0027s a picture of this pointwise limit."},{"Start":"03:44.885 ","End":"03:47.030","Text":"I should have highlighted this part,"},{"Start":"03:47.030 ","End":"03:48.949","Text":"but this is part of the graph."},{"Start":"03:48.949 ","End":"03:50.960","Text":"It starts out at 0,"},{"Start":"03:50.960 ","End":"03:53.480","Text":"stays 0 and at 1,"},{"Start":"03:53.480 ","End":"03:56.210","Text":"it suddenly jumps and gets the value 1."},{"Start":"03:56.210 ","End":"03:57.920","Text":"Just for interest\u0027s sake,"},{"Start":"03:57.920 ","End":"04:01.820","Text":"here\u0027s a graph of the different f_ns."},{"Start":"04:01.820 ","End":"04:03.290","Text":"Specifically f1, f2,"},{"Start":"04:03.290 ","End":"04:05.675","Text":"f3, f4, f12,"},{"Start":"04:05.675 ","End":"04:11.000","Text":"and you can see it squashes and squashes until eventually,"},{"Start":"04:11.000 ","End":"04:12.320","Text":"it reaches this shape,"},{"Start":"04:12.320 ","End":"04:15.065","Text":"which jumps suddenly to 1."},{"Start":"04:15.065 ","End":"04:18.450","Text":"That concludes this clip."}],"ID":28664},{"Watched":false,"Name":"Tutorial - Uniform Convergence","Duration":"3m 37s","ChapterTopicVideoID":27510,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.590","Text":"We just talked about pointwise convergence and now let\u0027s talk about"},{"Start":"00:04.590 ","End":"00:09.255","Text":"another kind of convergence called uniform convergence of functions."},{"Start":"00:09.255 ","End":"00:14.130","Text":"We saw in our example earlier that the functions"},{"Start":"00:14.130 ","End":"00:18.980","Text":"f_n(x) equals x^n are all continuous on the interval 0,"},{"Start":"00:18.980 ","End":"00:24.585","Text":"1 and this sequence of functions converges to a function f pointwise,"},{"Start":"00:24.585 ","End":"00:27.840","Text":"where F is the following function."},{"Start":"00:27.840 ","End":"00:32.325","Text":"It\u0027s mostly 0 except at 1 where it suddenly jumps to 1."},{"Start":"00:32.325 ","End":"00:34.580","Text":"But this is not continuous on 0,"},{"Start":"00:34.580 ","End":"00:37.695","Text":"1 because it has this jump, this step."},{"Start":"00:37.695 ","End":"00:40.455","Text":"We\u0027re going to define uniform convergence."},{"Start":"00:40.455 ","End":"00:43.160","Text":"One of the advantages is that if"},{"Start":"00:43.160 ","End":"00:51.365","Text":"the sequence f_n are all continuous and f_n converges to f uniformly,"},{"Start":"00:51.365 ","End":"00:55.895","Text":"then we\u0027ll know that the limit is also continuous,"},{"Start":"00:55.895 ","End":"00:59.165","Text":"which doesn\u0027t happen with pointwise convergence."},{"Start":"00:59.165 ","End":"01:01.145","Text":"Now let\u0027s define it."},{"Start":"01:01.145 ","End":"01:04.560","Text":"Let\u0027s f_n(x), where n goes from 1, 2, 3, etc."},{"Start":"01:04.560 ","End":"01:09.835","Text":"Be a sequence of real or complex functions defined on some domain I."},{"Start":"01:09.835 ","End":"01:16.160","Text":"We say that the sequence f_n converges uniformly on I to f,"},{"Start":"01:16.160 ","End":"01:18.365","Text":"which is a function also defined on I,"},{"Start":"01:18.365 ","End":"01:21.430","Text":"if the following condition holds."},{"Start":"01:21.430 ","End":"01:26.374","Text":"The supernum on the domain I of the absolute value of the difference"},{"Start":"01:26.374 ","End":"01:31.885","Text":"f_n minus f converges to 0 as then goes to infinity."},{"Start":"01:31.885 ","End":"01:33.590","Text":"If this condition occurs,"},{"Start":"01:33.590 ","End":"01:36.620","Text":"we write that the limit as n goes to infinity of f_n equals"},{"Start":"01:36.620 ","End":"01:40.405","Text":"f uniformly or one of these arrows."},{"Start":"01:40.405 ","End":"01:44.255","Text":"The function f is said to be the uniform limit of f_n."},{"Start":"01:44.255 ","End":"01:47.225","Text":"Now a theorem which we won\u0027t prove here,"},{"Start":"01:47.225 ","End":"01:51.710","Text":"if f_n is a sequence of continuous functions on a domain"},{"Start":"01:51.710 ","End":"01:56.255","Text":"I and f_n converges to f uniformly on I,"},{"Start":"01:56.255 ","End":"02:00.635","Text":"then the limit f is also continuous on I."},{"Start":"02:00.635 ","End":"02:02.690","Text":"Now a couple of remarks."},{"Start":"02:02.690 ","End":"02:08.285","Text":"First of all, if f_n converges to f uniformly,"},{"Start":"02:08.285 ","End":"02:11.270","Text":"then it also converges pointwise."},{"Start":"02:11.270 ","End":"02:15.005","Text":"The reason is that if the supernum goes to 0,"},{"Start":"02:15.005 ","End":"02:17.765","Text":"then if I choose a particular x,"},{"Start":"02:17.765 ","End":"02:20.720","Text":"this absolute value will be less than or equal to the supernum,"},{"Start":"02:20.720 ","End":"02:22.640","Text":"so it also goes to 0."},{"Start":"02:22.640 ","End":"02:25.730","Text":"The second remark is that a sequence of"},{"Start":"02:25.730 ","End":"02:30.800","Text":"discontinuous functions can converge uniformly to a continuous function."},{"Start":"02:30.800 ","End":"02:33.410","Text":"It can converge to a discontinuous function,"},{"Start":"02:33.410 ","End":"02:35.885","Text":"you just can\u0027t say anything about it."},{"Start":"02:35.885 ","End":"02:39.080","Text":"The third remark is that a sequence of"},{"Start":"02:39.080 ","End":"02:44.285","Text":"continuous functions could converge pointwise to a continuous function,"},{"Start":"02:44.285 ","End":"02:46.640","Text":"but not uniformly to that function."},{"Start":"02:46.640 ","End":"02:48.965","Text":"It\u0027s possible there are examples."},{"Start":"02:48.965 ","End":"02:51.830","Text":"Now let\u0027s do an example exercise."},{"Start":"02:51.830 ","End":"03:01.190","Text":"Check the uniform convergence of the sequence f_n(x) equals x^n on the interval from 0-1."},{"Start":"03:01.190 ","End":"03:05.360","Text":"If f_n converges uniformly, then as we mentioned,"},{"Start":"03:05.360 ","End":"03:08.930","Text":"it also converges pointwise to the same limit and we showed"},{"Start":"03:08.930 ","End":"03:13.645","Text":"this pointwise limit to be the following step function."},{"Start":"03:13.645 ","End":"03:19.100","Text":"F_n converges uniformly to this F. But by the theorem,"},{"Start":"03:19.100 ","End":"03:26.224","Text":"f should be continuous and our f is not continuous, violating the theorem."},{"Start":"03:26.224 ","End":"03:29.405","Text":"It\u0027s a contradiction and so therefore,"},{"Start":"03:29.405 ","End":"03:34.425","Text":"f_n does not converge uniformly on 0, 1."},{"Start":"03:34.425 ","End":"03:38.020","Text":"With that, we conclude this clip."}],"ID":28665},{"Watched":false,"Name":"Tutorial - Convergence in Norm","Duration":"3m 56s","ChapterTopicVideoID":27511,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.270","Text":"Now we come to another kind of convergence."},{"Start":"00:03.270 ","End":"00:06.090","Text":"Up to now, we had pointwise and uniform,"},{"Start":"00:06.090 ","End":"00:08.685","Text":"and now we have convergence in norm."},{"Start":"00:08.685 ","End":"00:11.055","Text":"This is good for normed spaces."},{"Start":"00:11.055 ","End":"00:13.065","Text":"Let V be a normed space,"},{"Start":"00:13.065 ","End":"00:14.385","Text":"and this is its norm."},{"Start":"00:14.385 ","End":"00:16.230","Text":"The dot is a placeholder."},{"Start":"00:16.230 ","End":"00:20.100","Text":"We say that a sequence f_n of vectors in V,"},{"Start":"00:20.100 ","End":"00:22.245","Text":"although the vectors here are functions,"},{"Start":"00:22.245 ","End":"00:25.125","Text":"converges in norm to f,"},{"Start":"00:25.125 ","End":"00:34.050","Text":"which is also in V if the limit is n goes to infinity of the norm of f_n minus f is 0."},{"Start":"00:34.050 ","End":"00:39.390","Text":"You think about it, it\u0027s similar to sequences of numbers xn"},{"Start":"00:39.390 ","End":"00:46.130","Text":"converges to x if the limit of the absolute value of xn minus x is 0."},{"Start":"00:46.130 ","End":"00:49.715","Text":"It\u0027s similar, but instead of absolute value, we have norm."},{"Start":"00:49.715 ","End":"00:53.020","Text":"Instead of numbers, we have vectors or functions."},{"Start":"00:53.020 ","End":"00:56.610","Text":"F is called the limit in norm of f_n."},{"Start":"00:56.610 ","End":"00:59.330","Text":"Here\u0027s an example exercise."},{"Start":"00:59.330 ","End":"01:03.620","Text":"We consider the space L^1 on the interval 0,1,"},{"Start":"01:03.620 ","End":"01:07.400","Text":"which is all the functions complex valued such"},{"Start":"01:07.400 ","End":"01:12.815","Text":"that the integral of the absolute value of f is finite."},{"Start":"01:12.815 ","End":"01:16.310","Text":"The norm is defined to be this integral,"},{"Start":"01:16.310 ","End":"01:18.410","Text":"which we know is finite."},{"Start":"01:18.410 ","End":"01:24.270","Text":"Our task is to check the convergence of f_n which is x^n,"},{"Start":"01:24.270 ","End":"01:26.267","Text":"like in our previous example,"},{"Start":"01:26.267 ","End":"01:29.920","Text":"to check the convergence in the L^1 norm."},{"Start":"01:29.920 ","End":"01:35.180","Text":"An obvious candidate for the limit in norm would be the pointwise limit."},{"Start":"01:35.180 ","End":"01:37.370","Text":"This is usually the case."},{"Start":"01:37.370 ","End":"01:39.575","Text":"We have to show 3 things."},{"Start":"01:39.575 ","End":"01:42.335","Text":"We have to show that the sequence f_n,"},{"Start":"01:42.335 ","End":"01:45.140","Text":"or each of them belongs to L^1."},{"Start":"01:45.140 ","End":"01:50.015","Text":"That the candidate belongs to L^1,"},{"Start":"01:50.015 ","End":"01:54.165","Text":"and then that this tends to this."},{"Start":"01:54.165 ","End":"01:58.375","Text":"First of all, why does f belong to L^1 on 0,1?"},{"Start":"01:58.375 ","End":"02:04.840","Text":"The integral of absolute value of f is the integral from 0-1(0)."},{"Start":"02:04.840 ","End":"02:07.160","Text":"Yes, this is not the 0 function,"},{"Start":"02:07.160 ","End":"02:10.815","Text":"but changing it at a single point doesn\u0027t change the integral."},{"Start":"02:10.815 ","End":"02:12.955","Text":"This integral is finite."},{"Start":"02:12.955 ","End":"02:15.955","Text":"Now, why does each f_n belong to L^1?"},{"Start":"02:15.955 ","End":"02:22.160","Text":"Well, could be that\u0027s because f_n is continuous on the interval and so the"},{"Start":"02:22.160 ","End":"02:28.360","Text":"integral will be bounded by this bound times the length of the interval, which is 1."},{"Start":"02:28.360 ","End":"02:30.070","Text":"Anyway, it\u0027s bounded."},{"Start":"02:30.070 ","End":"02:33.620","Text":"The last part is that f_n tends to f. In other words,"},{"Start":"02:33.620 ","End":"02:39.190","Text":"that this limit of f_n minus f in the norm is 0."},{"Start":"02:39.190 ","End":"02:41.830","Text":"This norm is the L^1 norm."},{"Start":"02:41.830 ","End":"02:49.779","Text":"We need to check that the integral of f_n minus f from 0-1 goes to 0."},{"Start":"02:49.779 ","End":"02:55.310","Text":"This is equal to the integral of x^n minus 0."},{"Start":"02:55.310 ","End":"02:58.015","Text":"We can drop the absolute value on the 0."},{"Start":"02:58.015 ","End":"03:04.600","Text":"This is x^n plus 1 over n plus 1 from 0-1 comes out 1 over n plus 1,"},{"Start":"03:04.600 ","End":"03:09.305","Text":"which certainly goes to 0 as n goes to infinity."},{"Start":"03:09.305 ","End":"03:11.375","Text":"That concludes the exercise."},{"Start":"03:11.375 ","End":"03:13.425","Text":"Now, remarks."},{"Start":"03:13.425 ","End":"03:16.725","Text":"Before we can say that f_n converges to f,"},{"Start":"03:16.725 ","End":"03:18.305","Text":"this is just a reminder."},{"Start":"03:18.305 ","End":"03:24.320","Text":"You\u0027ve got to have that each f_n belongs to the space and that f belongs to the space."},{"Start":"03:24.320 ","End":"03:26.409","Text":"Otherwise, it\u0027s meaningless."},{"Start":"03:26.409 ","End":"03:35.030","Text":"Secondly, if the limit of f_n minus f in the space V is 0,"},{"Start":"03:35.030 ","End":"03:40.010","Text":"then the norm of f_n converges to the norm of f,"},{"Start":"03:40.010 ","End":"03:42.530","Text":"we\u0027ll prove this in the exercises."},{"Start":"03:42.530 ","End":"03:49.220","Text":"One conclusion from this is that if we show that the norm of f_n goes to infinity,"},{"Start":"03:49.220 ","End":"03:53.310","Text":"then the sequence f_n doesn\u0027t converge in norm."},{"Start":"03:53.310 ","End":"03:57.130","Text":"With that, we conclude this clip."}],"ID":28666},{"Watched":false,"Name":"Exercise 1 - Part a","Duration":"3m 4s","ChapterTopicVideoID":27484,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.030","Text":"In this exercise, f_n is a sequence of functions that are continuous on"},{"Start":"00:06.030 ","End":"00:13.420","Text":"the closed interval from 0-10 and the function is given in pieces as shown."},{"Start":"00:13.420 ","End":"00:18.620","Text":"Note that it\u0027s defined separately on 3 closed interval from"},{"Start":"00:18.620 ","End":"00:23.750","Text":"0-1 over n for 1 over n to 2 over n and from 2 over n up to 10."},{"Start":"00:23.750 ","End":"00:24.935","Text":"There is an overlap,"},{"Start":"00:24.935 ","End":"00:27.790","Text":"but on the overlap, these are equal."},{"Start":"00:27.790 ","End":"00:29.360","Text":"It is continuous."},{"Start":"00:29.360 ","End":"00:31.430","Text":"You can check the computations."},{"Start":"00:31.430 ","End":"00:39.655","Text":"In part a, we have to check if f_n converges pointwise on the interval."},{"Start":"00:39.655 ","End":"00:42.805","Text":"Then in parts b and c and d,"},{"Start":"00:42.805 ","End":"00:45.050","Text":"we have to check different convergence;"},{"Start":"00:45.050 ","End":"00:53.260","Text":"uniform convergence, convergence in the norm of L^1 0,10 and convergence in L^2 0,10."},{"Start":"00:53.260 ","End":"00:57.530","Text":"Certain ones imply others for example,"},{"Start":"00:57.530 ","End":"01:00.410","Text":"if it converges uniformly,"},{"Start":"01:00.410 ","End":"01:02.660","Text":"then it converges pointwise."},{"Start":"01:02.660 ","End":"01:04.490","Text":"In fact, if it converges uniformly,"},{"Start":"01:04.490 ","End":"01:11.420","Text":"then it also converges in L^1 and L^2 because it\u0027s a compact interval."},{"Start":"01:11.420 ","End":"01:13.850","Text":"Also if it converges in L^2,"},{"Start":"01:13.850 ","End":"01:15.965","Text":"then it converges in L^1."},{"Start":"01:15.965 ","End":"01:19.045","Text":"Well, let\u0027s see. Let\u0027s start off with part a."},{"Start":"01:19.045 ","End":"01:24.455","Text":"In fact, the functions f_n converge pointwise to 0,"},{"Start":"01:24.455 ","End":"01:27.290","Text":"or you could say to the function f=0."},{"Start":"01:27.290 ","End":"01:34.265","Text":"The thing with pointwise convergence is that we pick any arbitrary x,"},{"Start":"01:34.265 ","End":"01:36.305","Text":"we have to show that for each x,"},{"Start":"01:36.305 ","End":"01:38.320","Text":"not depending on n,"},{"Start":"01:38.320 ","End":"01:43.510","Text":"that f_n(x) converges to f(x) as n goes to infinity,"},{"Start":"01:43.510 ","End":"01:47.005","Text":"and we\u0027ll split it up into 2 cases."},{"Start":"01:47.005 ","End":"01:51.775","Text":"X not equal to 0 and x=0 so if x is not 0,"},{"Start":"01:51.775 ","End":"01:54.910","Text":"it\u0027s from 0-10 excluding the 0."},{"Start":"01:54.910 ","End":"01:57.385","Text":"Since x is bigger than 0,"},{"Start":"01:57.385 ","End":"02:02.950","Text":"we can always choose n such that 2 over n is less than x,"},{"Start":"02:02.950 ","End":"02:07.510","Text":"meaning that all this part is to the left of x."},{"Start":"02:07.510 ","End":"02:11.770","Text":"You can do it by choosing n bigger than 2 over x, for example."},{"Start":"02:11.770 ","End":"02:13.885","Text":"Now what that means,"},{"Start":"02:13.885 ","End":"02:17.385","Text":"if 2 over n is less than x,"},{"Start":"02:17.385 ","End":"02:21.695","Text":"is that f_n(x) is 0 for this"},{"Start":"02:21.695 ","End":"02:28.020","Text":"n. F_n(x) is 0 if n is bigger than 2 over x,"},{"Start":"02:28.020 ","End":"02:30.385","Text":"which means from a certain point onwards,"},{"Start":"02:30.385 ","End":"02:32.769","Text":"all the f_n\u0027s are 0."},{"Start":"02:32.769 ","End":"02:35.680","Text":"If this is true from a certain point onwards,"},{"Start":"02:35.680 ","End":"02:38.365","Text":"then the limit is 0."},{"Start":"02:38.365 ","End":"02:41.080","Text":"That\u0027s the case where x is bigger than 0."},{"Start":"02:41.080 ","End":"02:44.755","Text":"Then in the case where x is equal to 0,"},{"Start":"02:44.755 ","End":"02:47.725","Text":"f_n for every n is 0."},{"Start":"02:47.725 ","End":"02:49.300","Text":"If we go back and look,"},{"Start":"02:49.300 ","End":"02:50.919","Text":"here in this interval,"},{"Start":"02:50.919 ","End":"02:54.565","Text":"if you plug in x=0, f_n(x) is 0."},{"Start":"02:54.565 ","End":"02:57.460","Text":"And so certainly the limit is 0."},{"Start":"02:57.460 ","End":"03:00.670","Text":"In any case, the limit of f n is 0."},{"Start":"03:00.670 ","End":"03:05.510","Text":"And so we have pointwise convergence, that\u0027s part a."}],"ID":28667},{"Watched":false,"Name":"Exercise 1 - Part b","Duration":"1m 43s","ChapterTopicVideoID":27485,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.250","Text":"Now we come to part b,"},{"Start":"00:02.250 ","End":"00:06.405","Text":"which asks about uniform convergence of this sequence."},{"Start":"00:06.405 ","End":"00:12.600","Text":"Now, if in part a we had concluded that f_n doesn\u0027t converge pointwise,"},{"Start":"00:12.600 ","End":"00:15.690","Text":"then we could say that it doesn\u0027t converge uniformly."},{"Start":"00:15.690 ","End":"00:18.830","Text":"However, we saw that it does converge pointwise,"},{"Start":"00:18.830 ","End":"00:23.760","Text":"so we can\u0027t say from that alone whether or not f_n converges uniformly,"},{"Start":"00:23.760 ","End":"00:26.730","Text":"we\u0027ll just have to check it directly."},{"Start":"00:26.730 ","End":"00:31.800","Text":"We will, in fact, show that it doesn\u0027t converge uniformly."},{"Start":"00:31.800 ","End":"00:36.630","Text":"Suppose it does, then the limit would be 0,"},{"Start":"00:36.630 ","End":"00:38.211","Text":"same as the pointwise,"},{"Start":"00:38.211 ","End":"00:42.225","Text":"and we\u0027ll show that f_n doesn\u0027t converge to 0 uniformly."},{"Start":"00:42.225 ","End":"00:47.390","Text":"The definition of uniform convergence is at the supremum of f_n"},{"Start":"00:47.390 ","End":"00:53.095","Text":"minus f in absolute value on the interval goes to 0 as n goes to infinity."},{"Start":"00:53.095 ","End":"00:58.510","Text":"Since f is 0, this is just the supremum of absolute value of f_n(x)."},{"Start":"00:58.510 ","End":"01:01.365","Text":"What we can do, instead of the supremum,"},{"Start":"01:01.365 ","End":"01:04.280","Text":"we could replace it by a particular point,"},{"Start":"01:04.280 ","End":"01:06.665","Text":"which is, in fact, the maximum."},{"Start":"01:06.665 ","End":"01:08.090","Text":"At 1 over n,"},{"Start":"01:08.090 ","End":"01:10.330","Text":"we get the largest value."},{"Start":"01:10.330 ","End":"01:15.155","Text":"The supremum, bigger or equal to any particular sample point."},{"Start":"01:15.155 ","End":"01:20.455","Text":"F_n(1) over n is square root of n. Just go back and look at the definition."},{"Start":"01:20.455 ","End":"01:23.580","Text":"Here on this interval or on this interval."},{"Start":"01:23.580 ","End":"01:25.920","Text":"Well, we\u0027ve already checked the computation here,"},{"Start":"01:25.920 ","End":"01:30.343","Text":"f_n(1 over n) is square root of n. This goes to infinity,"},{"Start":"01:30.343 ","End":"01:31.900","Text":"and this is bigger or equal to this,"},{"Start":"01:31.900 ","End":"01:38.450","Text":"so this supremum of the difference also goes to infinity and ensure it doesn\u0027t go to 0,"},{"Start":"01:38.450 ","End":"01:41.375","Text":"so we don\u0027t have uniform convergence."},{"Start":"01:41.375 ","End":"01:44.110","Text":"That does part b."}],"ID":28668},{"Watched":false,"Name":"Exercise 1 - Part c","Duration":"2m 26s","ChapterTopicVideoID":27486,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.530","Text":"Now we come to part c where we have to check if the Sequence f_n"},{"Start":"00:04.530 ","End":"00:08.895","Text":"converges in the norm of L^1 on the interval 0,10."},{"Start":"00:08.895 ","End":"00:11.504","Text":"I\u0027ll remind you what the norm is."},{"Start":"00:11.504 ","End":"00:17.205","Text":"The norm on this space is the integral from 0 to 10 of the absolute value."},{"Start":"00:17.205 ","End":"00:20.295","Text":"Now to show this, we have to actually show 3 things."},{"Start":"00:20.295 ","End":"00:25.425","Text":"We have to show that each of the f_n belongs to the space L^1."},{"Start":"00:25.425 ","End":"00:27.375","Text":"We have to show that the limit,"},{"Start":"00:27.375 ","End":"00:29.910","Text":"in this case, f equals 0,"},{"Start":"00:29.910 ","End":"00:34.050","Text":"will be the limit up to show that belongs here and we have to show that"},{"Start":"00:34.050 ","End":"00:39.160","Text":"the norm of the difference of f_n minus f goes to 0."},{"Start":"00:39.160 ","End":"00:43.730","Text":"Let\u0027s do the first thing to show that each of this f _n is in this space."},{"Start":"00:43.730 ","End":"00:48.770","Text":"What we have to show is that the norm of each 1 is less than infinity."},{"Start":"00:48.770 ","End":"00:54.575","Text":"Now, since f_n is continuous on this closed interval,"},{"Start":"00:54.575 ","End":"00:57.850","Text":"and the absolute value of f_n is bounded,"},{"Start":"00:57.850 ","End":"00:59.975","Text":"and since it\u0027s bounded,"},{"Start":"00:59.975 ","End":"01:06.275","Text":"the integral of a bounded function on a finite interval has to be less than infinity."},{"Start":"01:06.275 ","End":"01:12.380","Text":"That\u0027s okay, and then the candidate for a limit will be f equals 0,"},{"Start":"01:12.380 ","End":"01:13.820","Text":"the pointwise limit,"},{"Start":"01:13.820 ","End":"01:18.710","Text":"and that\u0027s also in the space because the 0 function is always in any vector space of"},{"Start":"01:18.710 ","End":"01:24.799","Text":"functions and so now we have to show the norm of the difference goes to 0."},{"Start":"01:24.799 ","End":"01:30.560","Text":"Let\u0027s see the normal of the difference is the integral of f_n minus f and absolute value,"},{"Start":"01:30.560 ","End":"01:32.255","Text":"but f is 0."},{"Start":"01:32.255 ","End":"01:35.990","Text":"It\u0027s just the integral of f_n is also non-negative,"},{"Start":"01:35.990 ","End":"01:37.880","Text":"so we can drop the absolute value,"},{"Start":"01:37.880 ","End":"01:43.485","Text":"and because f_n is 0,1, 2^n to 10."},{"Start":"01:43.485 ","End":"01:50.395","Text":"We can take this integral just from 0 to 2 over n because all the rest of it is 0,"},{"Start":"01:50.395 ","End":"01:52.595","Text":"and to compute this integral,"},{"Start":"01:52.595 ","End":"01:55.760","Text":"Let\u0027s go back and look at the function."},{"Start":"01:55.760 ","End":"01:58.750","Text":"It\u0027s the area under the curve which is a triangle,"},{"Start":"01:58.750 ","End":"02:02.150","Text":"and we know the formula for the area of a triangle,"},{"Start":"02:02.150 ","End":"02:04.870","Text":"a half base times height."},{"Start":"02:04.870 ","End":"02:08.285","Text":"Half base times height is a half."},{"Start":"02:08.285 ","End":"02:11.750","Text":"The base is 2 over n. The height is square root of"},{"Start":"02:11.750 ","End":"02:15.545","Text":"n. That comes out to be 1 over the square root of n,"},{"Start":"02:15.545 ","End":"02:18.259","Text":"and that does indeed tend to 0."},{"Start":"02:18.259 ","End":"02:20.175","Text":"The answer is yes,"},{"Start":"02:20.175 ","End":"02:23.595","Text":"f_n converges to f in the norm of L^1,"},{"Start":"02:23.595 ","End":"02:26.560","Text":"and that concludes part c."}],"ID":28669},{"Watched":false,"Name":"Exercise 1 - Part d","Duration":"2m 44s","ChapterTopicVideoID":27487,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"Now we come to the last part d,"},{"Start":"00:02.790 ","End":"00:06.375","Text":"where we have to check if a sequence f_n converges in L^2."},{"Start":"00:06.375 ","End":"00:11.340","Text":"It will be very similar to part c. Just different norm."},{"Start":"00:11.340 ","End":"00:13.665","Text":"We\u0027ll go a bit quicker this time."},{"Start":"00:13.665 ","End":"00:17.280","Text":"Here\u0027s a reminder of what the norm is in L^2."},{"Start":"00:17.280 ","End":"00:20.100","Text":"It\u0027s the integral of the absolute value squared."},{"Start":"00:20.100 ","End":"00:22.275","Text":"But then we take the square root."},{"Start":"00:22.275 ","End":"00:26.040","Text":"As before, we need to check that the f_n belongs to the space"},{"Start":"00:26.040 ","End":"00:31.455","Text":"L^2 on 0,10 to show that norm is less than infinity."},{"Start":"00:31.455 ","End":"00:34.200","Text":"Like before, f_n is continuous,"},{"Start":"00:34.200 ","End":"00:39.930","Text":"so the absolute value of f_n^2 is bounded and non-negative"},{"Start":"00:39.930 ","End":"00:47.535","Text":"and so the norm squared will be less than infinity."},{"Start":"00:47.535 ","End":"00:52.650","Text":"This time f_n does not converge in L^1 it did converge,"},{"Start":"00:52.650 ","End":"00:56.675","Text":"in L^2, it turns out not to converge, but similar calculation."},{"Start":"00:56.675 ","End":"00:59.285","Text":"The limit if it converge would be 0,"},{"Start":"00:59.285 ","End":"01:02.165","Text":"just like in L^1."},{"Start":"01:02.165 ","End":"01:06.830","Text":"We already showed previously that if f_n converges to f and L^2,"},{"Start":"01:06.830 ","End":"01:10.880","Text":"then it converges in L^1 to the same limit."},{"Start":"01:10.880 ","End":"01:14.510","Text":"We take the norm squared easier to deal with than having"},{"Start":"01:14.510 ","End":"01:19.520","Text":"the square root is this integral of absolute value of f_n^2."},{"Start":"01:19.520 ","End":"01:22.105","Text":"Now the integrand is non-negative."},{"Start":"01:22.105 ","End":"01:26.165","Text":"If we take just the integral from 0-1 over n,"},{"Start":"01:26.165 ","End":"01:30.475","Text":"then it will only be less than or equal to the integral from 0-10."},{"Start":"01:30.475 ","End":"01:32.190","Text":"You want to show this as infinity,"},{"Start":"01:32.190 ","End":"01:33.975","Text":"so we want it bounded from below."},{"Start":"01:33.975 ","End":"01:39.440","Text":"It\u0027s bounded from below by the integral from 0-1 over n. Recall"},{"Start":"01:39.440 ","End":"01:46.205","Text":"that f_n(x) is n root n times x on this interval from 0-1 over n,"},{"Start":"01:46.205 ","End":"01:54.155","Text":"so f_n^2 is n root n^2 is n^3 and x^2 is x^2."},{"Start":"01:54.155 ","End":"01:55.730","Text":"This is what we have."},{"Start":"01:55.730 ","End":"02:04.980","Text":"Now, n^3 comes out in front and the integral of x^2 is 1/3x^3 and we plug in x equals"},{"Start":"02:04.980 ","End":"02:09.470","Text":"1 over n and then 0 and subtract when 0 gives nothing but x^3"},{"Start":"02:09.470 ","End":"02:15.043","Text":"over 3 when x is 1 over n comes out 1/3n^3 altogether."},{"Start":"02:15.043 ","End":"02:18.185","Text":"N^3 with n^3 cancels and we get 1/3."},{"Start":"02:18.185 ","End":"02:22.040","Text":"This integral is bigger or equal to 1/3,"},{"Start":"02:22.040 ","End":"02:25.490","Text":"so the absolute value in L^2 is bigger or equal"},{"Start":"02:25.490 ","End":"02:29.240","Text":"to square root of a 1/3 because we had squared here, it doesn\u0027t matter."},{"Start":"02:29.240 ","End":"02:32.645","Text":"It\u0027s still bigger or equal to something that\u0027s bigger than 0."},{"Start":"02:32.645 ","End":"02:36.410","Text":"It never tends to 0 as n goes to infinity,"},{"Start":"02:36.410 ","End":"02:41.090","Text":"and so f_n does not converge to f in the norm of L^2."},{"Start":"02:41.090 ","End":"02:44.760","Text":"That concludes part d, and we\u0027re done."}],"ID":28670},{"Watched":false,"Name":"Exercise 2 - Part a","Duration":"2m 9s","ChapterTopicVideoID":27488,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.550","Text":"In this exercise, we\u0027re given"},{"Start":"00:02.550 ","End":"00:08.700","Text":"a sequence f_n of continuous functions on the closed interval 0,1,"},{"Start":"00:08.700 ","End":"00:10.875","Text":"as in this formula."},{"Start":"00:10.875 ","End":"00:12.990","Text":"Like in the previous exercise,"},{"Start":"00:12.990 ","End":"00:17.230","Text":"we have 4 questions to check convergence point wise"},{"Start":"00:17.230 ","End":"00:23.199","Text":"uniformly in L^1 of the integral and in L^2 of the interval."},{"Start":"00:23.199 ","End":"00:26.010","Text":"We\u0027ll start with part a."},{"Start":"00:26.010 ","End":"00:32.055","Text":"It turns out that it does converge point wise to the 0 function on the interval."},{"Start":"00:32.055 ","End":"00:35.500","Text":"In fact, I\u0027ll show you a picture of this."},{"Start":"00:35.500 ","End":"00:38.310","Text":"If x is 0 or 1,"},{"Start":"00:38.310 ","End":"00:40.770","Text":"we\u0027ll take these 2 special cases."},{"Start":"00:40.770 ","End":"00:46.550","Text":"Then f_n(x) is 0 for all n. If you go back and look at the formula."},{"Start":"00:46.550 ","End":"00:48.515","Text":"If x is 0,"},{"Start":"00:48.515 ","End":"00:50.900","Text":"then x^n is 0 and if x is 1,"},{"Start":"00:50.900 ","End":"00:54.845","Text":"then 1 minus x is 0 so for both those cases, we\u0027re okay."},{"Start":"00:54.845 ","End":"00:58.096","Text":"Now what happens if x is strictly between 0 and 1?"},{"Start":"00:58.096 ","End":"01:01.330","Text":"We\u0027ll show that f_n(x) tends to 0."},{"Start":"01:01.330 ","End":"01:05.480","Text":"In this limit, n goes to infinity and n doesn\u0027t depend on x so"},{"Start":"01:05.480 ","End":"01:09.965","Text":"we could take the 1 minus x in front of the limit."},{"Start":"01:09.965 ","End":"01:15.500","Text":"Then what we can do is let m be equal to 1"},{"Start":"01:15.500 ","End":"01:21.065","Text":"over x and 1 over x is bigger than 1 because x is less than 1."},{"Start":"01:21.065 ","End":"01:24.770","Text":"This is equal to limit of x^n."},{"Start":"01:24.770 ","End":"01:28.040","Text":"We can put m^n in the denominator."},{"Start":"01:28.040 ","End":"01:31.340","Text":"You could do it by L\u0027Hopital\u0027s rule."},{"Start":"01:31.340 ","End":"01:34.430","Text":"If you think of n as a continuous variable,"},{"Start":"01:34.430 ","End":"01:39.075","Text":"you could differentiate numerator and denominator and get that"},{"Start":"01:39.075 ","End":"01:47.300","Text":"the denominator goes to infinity and the numerator is 1, the derivative."},{"Start":"01:47.300 ","End":"01:49.145","Text":"I could use the ratio test."},{"Start":"01:49.145 ","End":"01:57.410","Text":"Anyway, this limit is 0 and so the limit as n goes to infinity is 0,"},{"Start":"01:57.410 ","End":"02:01.340","Text":"because 1 minus x is just some number not dependent on n"},{"Start":"02:01.340 ","End":"02:06.440","Text":"so times 0 is still 0 and that 0 is f(x)."},{"Start":"02:06.440 ","End":"02:10.140","Text":"This concludes part a."}],"ID":28671},{"Watched":false,"Name":"Exercise 2 - Part b","Duration":"2m 37s","ChapterTopicVideoID":27489,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.640","Text":"Now we come to part b."},{"Start":"00:02.640 ","End":"00:06.315","Text":"If in part a, we had the answer no,"},{"Start":"00:06.315 ","End":"00:10.260","Text":"then we could have concluded in b that the answer is also no because"},{"Start":"00:10.260 ","End":"00:14.220","Text":"uniform convergence implies pointwise convergence."},{"Start":"00:14.220 ","End":"00:15.930","Text":"But we got the answer yes here,"},{"Start":"00:15.930 ","End":"00:19.335","Text":"so we don\u0027t know here so we have to check it."},{"Start":"00:19.335 ","End":"00:25.770","Text":"In fact, the answer I know is that it does not converge uniformly on 0,1."},{"Start":"00:25.770 ","End":"00:27.090","Text":"Let\u0027s show this."},{"Start":"00:27.090 ","End":"00:29.609","Text":"But if it did converge uniformly,"},{"Start":"00:29.609 ","End":"00:36.480","Text":"the limit would also be f=0 and we\u0027ll show that f_n does not converge to 0 uniformly."},{"Start":"00:36.480 ","End":"00:42.330","Text":"A uniform convergence means that the supremum of the difference goes to 0,"},{"Start":"00:42.330 ","End":"00:43.610","Text":"then goes to infinity."},{"Start":"00:43.610 ","End":"00:45.515","Text":"Let\u0027s see what this is."},{"Start":"00:45.515 ","End":"00:47.930","Text":"This is equal to the supremum."},{"Start":"00:47.930 ","End":"00:51.425","Text":"We don\u0027t need the absolute value because it\u0027s non-negative."},{"Start":"00:51.425 ","End":"00:54.010","Text":"We have n(1 minus x)x_n."},{"Start":"00:54.010 ","End":"00:57.505","Text":"We can take n out because the supremum is over x."},{"Start":"00:57.505 ","End":"00:59.250","Text":"Because it\u0027s continuous, we can talk about"},{"Start":"00:59.250 ","End":"01:03.665","Text":"the maximum and we can find the maximum using calculus."},{"Start":"01:03.665 ","End":"01:07.505","Text":"We can take the derivative and find out where that is 0."},{"Start":"01:07.505 ","End":"01:09.080","Text":"Here\u0027s the derivative,"},{"Start":"01:09.080 ","End":"01:17.870","Text":"let it equal 0 and we can take x_n minus 1 outside the brackets and get this is 0."},{"Start":"01:17.870 ","End":"01:25.720","Text":"A bit of algebra and we get that x=n over n pus 1."},{"Start":"01:26.000 ","End":"01:28.815","Text":"Here\u0027s a picture."},{"Start":"01:28.815 ","End":"01:31.680","Text":"This is the point at which we get the maximum."},{"Start":"01:31.680 ","End":"01:35.135","Text":"Let\u0027s see what the maximum value is."},{"Start":"01:35.135 ","End":"01:38.810","Text":"We know it\u0027s a maximum because we have a non-negative function"},{"Start":"01:38.810 ","End":"01:42.469","Text":"that\u0027s zero here and here so it can\u0027t be a minimum."},{"Start":"01:42.469 ","End":"01:44.390","Text":"It can\u0027t be going down and up."},{"Start":"01:44.390 ","End":"01:48.630","Text":"Or you could use second derivative or advanced techniques,"},{"Start":"01:48.630 ","End":"01:51.680","Text":"but it\u0027s pretty clear that under these conditions it"},{"Start":"01:51.680 ","End":"01:54.800","Text":"has to be a maximum and not a minimum or an inflection."},{"Start":"01:54.800 ","End":"01:56.060","Text":"Let\u0027s see."},{"Start":"01:56.060 ","End":"02:00.320","Text":"The supremum is n times the maximum of this,"},{"Start":"02:00.320 ","End":"02:07.100","Text":"which is like we said when x=n over n plus 1 and just do,"},{"Start":"02:07.100 ","End":"02:11.330","Text":"again a bit of algebra and we get this expression."},{"Start":"02:11.330 ","End":"02:15.800","Text":"Now we know that this tends to e as n goes to infinity."},{"Start":"02:15.800 ","End":"02:17.250","Text":"It\u0027s a well-known limit."},{"Start":"02:17.250 ","End":"02:19.000","Text":"This goes to 1,"},{"Start":"02:19.000 ","End":"02:26.490","Text":"so 1 times 1 over e. It tends to 1 over e. This is not"},{"Start":"02:26.490 ","End":"02:35.255","Text":"0 so the supremum does not tend to 0 and so there is no uniform convergence of f_n to f,"},{"Start":"02:35.255 ","End":"02:38.010","Text":"and that concludes part b."}],"ID":28672},{"Watched":false,"Name":"Exercise 2 - Part c","Duration":"2m 13s","ChapterTopicVideoID":27490,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:09.210","Text":"Now we come to part c. We have to check if f_n converges in the norm of L^1, 0,1."},{"Start":"00:09.210 ","End":"00:13.485","Text":"Here\u0027s a reminder of what the norm is in this space,"},{"Start":"00:13.485 ","End":"00:16.455","Text":"it\u0027s the integral of the absolute value."},{"Start":"00:16.455 ","End":"00:19.260","Text":"What we have to check first of all,"},{"Start":"00:19.260 ","End":"00:22.980","Text":"is that each f_n belongs to the space."},{"Start":"00:22.980 ","End":"00:27.300","Text":"In other words that this norm is less than infinity. Let\u0027s do that."},{"Start":"00:27.300 ","End":"00:31.520","Text":"Now the integral of f_n is the integral of this drop,"},{"Start":"00:31.520 ","End":"00:33.950","Text":"the absolute value because it\u0027s non-negative."},{"Start":"00:33.950 ","End":"00:38.975","Text":"Take the n upfront because it doesn\u0027t depend on x. F is straightforward integral."},{"Start":"00:38.975 ","End":"00:44.750","Text":"This is x^n plus 1 over n plus 1 minus x^n plus 2 over n plus 2."},{"Start":"00:44.750 ","End":"00:46.685","Text":"If you plug in 1,"},{"Start":"00:46.685 ","End":"00:48.335","Text":"we get this,"},{"Start":"00:48.335 ","End":"00:50.240","Text":"plug in 0, it\u0027s 0."},{"Start":"00:50.240 ","End":"00:52.945","Text":"This is what the integral is."},{"Start":"00:52.945 ","End":"00:57.050","Text":"We can do a bit of algebra to simplify it and won\u0027t go into the details."},{"Start":"00:57.050 ","End":"00:59.660","Text":"We end up with n over n plus 1, n plus 2."},{"Start":"00:59.660 ","End":"01:03.710","Text":"The point is that this is some number for any given n. This is a number"},{"Start":"01:03.710 ","End":"01:07.425","Text":"which is a finite number less than infinity,"},{"Start":"01:07.425 ","End":"01:09.220","Text":"so we\u0027re okay there."},{"Start":"01:09.220 ","End":"01:13.790","Text":"You could save the calculations by noting that f is continuous."},{"Start":"01:13.790 ","End":"01:16.100","Text":"Absolute value of f is continuous,"},{"Start":"01:16.100 ","End":"01:19.205","Text":"so it\u0027s bounded on the closed interval 0,1."},{"Start":"01:19.205 ","End":"01:24.860","Text":"The integral of a bounded function on A compact interval is certainly less than infinity."},{"Start":"01:24.860 ","End":"01:26.975","Text":"We could do that also."},{"Start":"01:26.975 ","End":"01:30.560","Text":"Now the candidate for a limit if there is one,"},{"Start":"01:30.560 ","End":"01:34.145","Text":"would be the pointwise limit, which is f=0."},{"Start":"01:34.145 ","End":"01:40.115","Text":"This certainly belongs to the space because the 0 function belongs to any vector space."},{"Start":"01:40.115 ","End":"01:46.510","Text":"It remains to show that the limit of f_n minus f is 0,"},{"Start":"01:46.510 ","End":"01:49.195","Text":"so just norm of f_n."},{"Start":"01:49.195 ","End":"01:51.020","Text":"Now by the computation above,"},{"Start":"01:51.020 ","End":"01:55.010","Text":"the norm of f_n is this expression here."},{"Start":"01:55.010 ","End":"01:58.610","Text":"This certainly goes to 0 when n goes to infinity because we have a degree"},{"Start":"01:58.610 ","End":"02:02.515","Text":"1 polynomial over degree 2 polynomial."},{"Start":"02:02.515 ","End":"02:05.180","Text":"Degree below is bigger than the degree above,"},{"Start":"02:05.180 ","End":"02:07.205","Text":"so it goes to 0."},{"Start":"02:07.205 ","End":"02:09.620","Text":"Yes, f_n does converge,"},{"Start":"02:09.620 ","End":"02:11.105","Text":"it converges to 0."},{"Start":"02:11.105 ","End":"02:13.980","Text":"That concludes part c."}],"ID":28673},{"Watched":false,"Name":"Exercise 2 - Part d","Duration":"3m 1s","ChapterTopicVideoID":27491,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"Now, we come to part d where we have to check"},{"Start":"00:03.030 ","End":"00:07.200","Text":"the convergence in L^2 of the interval 0,1."},{"Start":"00:07.200 ","End":"00:11.880","Text":"Here\u0027s a reminder of what the norm of L^2 is in part c,"},{"Start":"00:11.880 ","End":"00:13.830","Text":"we got the answer yes."},{"Start":"00:13.830 ","End":"00:18.075","Text":"That\u0027s not useful for part d. If we had an answer no here,"},{"Start":"00:18.075 ","End":"00:20.475","Text":"then the answer here would also be no."},{"Start":"00:20.475 ","End":"00:26.870","Text":"But L^2 convergence implies L^1 convergence so we can\u0027t say."},{"Start":"00:26.870 ","End":"00:30.530","Text":"Anyway, it turns out that the answer is yes, as we shall see."},{"Start":"00:30.530 ","End":"00:35.540","Text":"We need to check first of all that these functions f_n really are in the space."},{"Start":"00:35.540 ","End":"00:38.740","Text":"In other words that their norm is less than infinity."},{"Start":"00:38.740 ","End":"00:40.605","Text":"Here\u0027s the computation."},{"Start":"00:40.605 ","End":"00:42.979","Text":"The norm, well, squared,"},{"Start":"00:42.979 ","End":"00:47.825","Text":"easier to work with the square of the norm, is this integral."},{"Start":"00:47.825 ","End":"00:51.020","Text":"That\u0027s equal to the following."},{"Start":"00:51.020 ","End":"00:52.910","Text":"Take the n squared out the brackets,"},{"Start":"00:52.910 ","End":"01:01.330","Text":"we have the following integral and open up the brackets and use a distributive law."},{"Start":"01:01.330 ","End":"01:02.850","Text":"This is what we have."},{"Start":"01:02.850 ","End":"01:06.260","Text":"Then the integral, these are just polynomials in"},{"Start":"01:06.260 ","End":"01:09.635","Text":"x is raised the power and divide by the power."},{"Start":"01:09.635 ","End":"01:12.935","Text":"Anyway, easily seen that this is the result,"},{"Start":"01:12.935 ","End":"01:16.395","Text":"we can simplify the middle term,"},{"Start":"01:16.395 ","End":"01:18.420","Text":"comes out to be 2 over 2n plus 2."},{"Start":"01:18.420 ","End":"01:21.385","Text":"We can write as 1 over n plus 1."},{"Start":"01:21.385 ","End":"01:24.440","Text":"Then using a bit of algebra,"},{"Start":"01:24.440 ","End":"01:27.605","Text":"putting a common denominator, we get this."},{"Start":"01:27.605 ","End":"01:30.050","Text":"Oops, I forgot the n squared here."},{"Start":"01:30.050 ","End":"01:32.015","Text":"No problem."},{"Start":"01:32.015 ","End":"01:34.385","Text":"Then simplifying that."},{"Start":"01:34.385 ","End":"01:36.020","Text":"Well, I won\u0027t bore you with the details."},{"Start":"01:36.020 ","End":"01:38.645","Text":"You can pause and take a look at the algebra."},{"Start":"01:38.645 ","End":"01:42.845","Text":"Finally we get that this numerator comes out to be 1."},{"Start":"01:42.845 ","End":"01:44.270","Text":"For example, n squared."},{"Start":"01:44.270 ","End":"01:47.830","Text":"We have 2 minus 4 plus 2 is 0,"},{"Start":"01:47.830 ","End":"01:50.925","Text":"and 5 minus 8 plus 3 is 0."},{"Start":"01:50.925 ","End":"01:53.955","Text":"We just have the 3 minus 3 plus 1."},{"Start":"01:53.955 ","End":"01:56.745","Text":"Numerators comes out to be 1,"},{"Start":"01:56.745 ","End":"01:59.765","Text":"1 with the n squared gives us n squared."},{"Start":"01:59.765 ","End":"02:02.450","Text":"Doesn\u0027t matter what the actual expression is."},{"Start":"02:02.450 ","End":"02:05.010","Text":"It\u0027s something finite."},{"Start":"02:05.030 ","End":"02:10.245","Text":"F_n really does belong to the space L^2."},{"Start":"02:10.245 ","End":"02:14.030","Text":"Or we could say, like we did in part c,"},{"Start":"02:14.030 ","End":"02:19.340","Text":"that f is continuous on the closed interval, etc."},{"Start":"02:19.340 ","End":"02:23.330","Text":"Now, a candidate for the limit would be the pointwise limit,"},{"Start":"02:23.330 ","End":"02:25.025","Text":"which is f=0,"},{"Start":"02:25.025 ","End":"02:28.955","Text":"and certainly 0 belongs to this space."},{"Start":"02:28.955 ","End":"02:37.490","Text":"New page. What remains to show is that the limit of the norm of the difference is 0."},{"Start":"02:37.490 ","End":"02:40.070","Text":"You can ignore this minus f because f is 0."},{"Start":"02:40.070 ","End":"02:44.810","Text":"You just have to show that the norm of f_n goes to 0."},{"Start":"02:44.810 ","End":"02:47.840","Text":"But we had the computation above for the norm of f_n,"},{"Start":"02:47.840 ","End":"02:49.520","Text":"it was this expression."},{"Start":"02:49.520 ","End":"02:53.615","Text":"There is a degree 2 polynomial over degree 3 polynomial."},{"Start":"02:53.615 ","End":"02:57.175","Text":"Certainly it goes to 0 when n goes to infinity."},{"Start":"02:57.175 ","End":"03:02.110","Text":"That concludes part d and this whole exercise."}],"ID":28674},{"Watched":false,"Name":"Exercise 3 - Part a","Duration":"1m 54s","ChapterTopicVideoID":27492,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.630","Text":"In this exercise, we have a sequence f_n of"},{"Start":"00:03.630 ","End":"00:08.190","Text":"functions from R to R and are defined as follows."},{"Start":"00:08.190 ","End":"00:14.565","Text":"1, if x is in the closed interval from n to n plus 1 over n and 0 otherwise,"},{"Start":"00:14.565 ","End":"00:22.210","Text":"then we have to check 4 different kinds of convergence pointwise uniform in L^1 and L^2."},{"Start":"00:22.210 ","End":"00:25.215","Text":"In contrast to previous exercises,"},{"Start":"00:25.215 ","End":"00:30.150","Text":"this is an infinite interval and in the case of an infinite interval,"},{"Start":"00:30.150 ","End":"00:32.955","Text":"usual implications don\u0027t hold."},{"Start":"00:32.955 ","End":"00:38.410","Text":"We can\u0027t say that uniform convergence implies convergence in L^1 and L^2,"},{"Start":"00:38.410 ","End":"00:42.985","Text":"and we can\u0027t say that converges in L^2 implies convergence in L^1."},{"Start":"00:42.985 ","End":"00:46.280","Text":"All that you can say that\u0027s always true is"},{"Start":"00:46.280 ","End":"00:50.195","Text":"that uniform convergence implies pointwise convergence."},{"Start":"00:50.195 ","End":"00:54.790","Text":"Anyway, let\u0027s start with part a where we have to check pointwise convergence."},{"Start":"00:54.790 ","End":"00:59.075","Text":"We show that it does converge pointwise to f equals 0."},{"Start":"00:59.075 ","End":"01:01.760","Text":"In other words, if for any x in r,"},{"Start":"01:01.760 ","End":"01:06.185","Text":"n of x tends to 0 as n goes to infinity."},{"Start":"01:06.185 ","End":"01:08.330","Text":"So given any x,"},{"Start":"01:08.330 ","End":"01:13.579","Text":"we can always find an integer N that\u0027s bigger or equal to x."},{"Start":"01:13.579 ","End":"01:17.080","Text":"Now, if n is bigger than N,"},{"Start":"01:17.080 ","End":"01:22.595","Text":"then x does not belong to the interval from n to n plus 1 over n,"},{"Start":"01:22.595 ","End":"01:26.420","Text":"because this whole interval is to the right of N,"},{"Start":"01:26.420 ","End":"01:28.760","Text":"which is to the right of x."},{"Start":"01:28.760 ","End":"01:32.540","Text":"In this interval, f_n of x is 0,"},{"Start":"01:32.540 ","End":"01:35.030","Text":"because here x does not belong to this."},{"Start":"01:35.030 ","End":"01:37.010","Text":"We\u0027re in the otherwise case."},{"Start":"01:37.010 ","End":"01:38.780","Text":"For enough n, in other words,"},{"Start":"01:38.780 ","End":"01:40.855","Text":"from a certainty n onwards,"},{"Start":"01:40.855 ","End":"01:43.385","Text":"we have that f_n of x is 0."},{"Start":"01:43.385 ","End":"01:49.250","Text":"That means that the limit of f_n of x is 0 as n goes to infinity and this is f of x."},{"Start":"01:49.250 ","End":"01:54.510","Text":"We\u0027ve just shown part a that f_n tends to f pointwise."}],"ID":28675},{"Watched":false,"Name":"Exercise 3 - Part b","Duration":"1m 23s","ChapterTopicVideoID":27493,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"Now we come to Part b."},{"Start":"00:02.160 ","End":"00:05.865","Text":"We\u0027re going to check uniform convergence."},{"Start":"00:05.865 ","End":"00:08.925","Text":"We know it converges pointwise,"},{"Start":"00:08.925 ","End":"00:12.630","Text":"but that doesn\u0027t tell us anything about whether or not it converges uniformly."},{"Start":"00:12.630 ","End":"00:14.760","Text":"It turns out that it doesn\u0027t."},{"Start":"00:14.760 ","End":"00:20.070","Text":"Intuitively, if it converged it\u0027d have to be to the 0 function,"},{"Start":"00:20.070 ","End":"00:27.660","Text":"but there\u0027s always a difference of 1 somewhere between our function and the 0 function."},{"Start":"00:27.660 ","End":"00:29.880","Text":"It\u0027s constantly a gap of 1."},{"Start":"00:29.880 ","End":"00:34.800","Text":"The answer would be no. Anyway, let\u0027s suppose it did converge uniformly."},{"Start":"00:34.800 ","End":"00:38.640","Text":"The limit would have to be the same as the pointwise limit of 0,"},{"Start":"00:38.640 ","End":"00:42.844","Text":"and we show that it isn\u0027t so by checking"},{"Start":"00:42.844 ","End":"00:50.570","Text":"the supremum of the absolute value of the difference of the two functions."},{"Start":"00:50.570 ","End":"00:52.140","Text":"f(x) is 0,"},{"Start":"00:52.140 ","End":"00:58.040","Text":"so we just have to check the absolute value of f_n(x) throughout the absolute value,"},{"Start":"00:58.040 ","End":"00:59.915","Text":"because it\u0027s bigger or equal to 0."},{"Start":"00:59.915 ","End":"01:07.690","Text":"It\u0027s bigger or equal to the value of f_n at any particular x. I can choose x=n."},{"Start":"01:07.690 ","End":"01:10.745","Text":"This has to only be bigger or equal to this supremum."},{"Start":"01:10.745 ","End":"01:14.180","Text":"But f_n(n)=1."},{"Start":"01:14.180 ","End":"01:17.150","Text":"This is always bigger or equal to 1,"},{"Start":"01:17.150 ","End":"01:20.375","Text":"so the limit cannot be 0."},{"Start":"01:20.375 ","End":"01:23.580","Text":"That concludes Part b."}],"ID":28676},{"Watched":false,"Name":"Exercise 3 - Part c","Duration":"1m 48s","ChapterTopicVideoID":27494,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"In part c,"},{"Start":"00:01.875 ","End":"00:06.885","Text":"we have to check if fn converges in the norm of L1."},{"Start":"00:06.885 ","End":"00:14.640","Text":"The first thing is to check that these functions fn really are in L1 of the real line."},{"Start":"00:14.640 ","End":"00:19.680","Text":"In other words, that the norm of f in L1 is finite,"},{"Start":"00:19.680 ","End":"00:22.680","Text":"but this is equal to the integral of"},{"Start":"00:22.680 ","End":"00:26.910","Text":"absolute value of fn from minus infinity to infinity."},{"Start":"00:26.910 ","End":"00:30.960","Text":"Well, let\u0027s see. Because fn and the absolute value of"},{"Start":"00:30.960 ","End":"00:36.713","Text":"fn are 0 outside of the interval from n to n plus 1,"},{"Start":"00:36.713 ","End":"00:39.275","Text":"we can reduce this interval that have"},{"Start":"00:39.275 ","End":"00:44.650","Text":"all the number line just to the part from n to n plus 1 over n,"},{"Start":"00:44.650 ","End":"00:48.785","Text":"where this is 1 and the absolute value of 1 is also 1."},{"Start":"00:48.785 ","End":"00:53.865","Text":"So this integral is n plus 1 over n minus n,"},{"Start":"00:53.865 ","End":"00:58.016","Text":"which is 1 over n. This is finite so"},{"Start":"00:58.016 ","End":"01:03.454","Text":"we\u0027re okay that fn really does belong to our space L1."},{"Start":"01:03.454 ","End":"01:06.395","Text":"Now we have to check the matter of convergence."},{"Start":"01:06.395 ","End":"01:09.755","Text":"Well, if it converges and it converges to 0."},{"Start":"01:09.755 ","End":"01:14.810","Text":"So let\u0027s check whether fn does converge to 0."},{"Start":"01:14.810 ","End":"01:16.325","Text":"We\u0027ll let f equal 0,"},{"Start":"01:16.325 ","End":"01:19.220","Text":"see if fn converges to f in L1."},{"Start":"01:19.220 ","End":"01:22.715","Text":"In other words, that the norm tends to 0."},{"Start":"01:22.715 ","End":"01:30.125","Text":"The norm of fn minus f is the same as the norm of fn, because f is 0."},{"Start":"01:30.125 ","End":"01:35.158","Text":"We already checked that this norm of fn is 1 over n,"},{"Start":"01:35.158 ","End":"01:41.885","Text":"and 1 over n does really tend to 0 as n goes to infinity so we\u0027re okay."},{"Start":"01:41.885 ","End":"01:44.660","Text":"The answer in part c is yes,"},{"Start":"01:44.660 ","End":"01:48.660","Text":"fn does converge to the 0 function."}],"ID":28677},{"Watched":false,"Name":"Exercise 3 - Part d","Duration":"1m 56s","ChapterTopicVideoID":27495,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.180","Text":"Now part d. So far we had the answer yes, no, yes."},{"Start":"00:06.180 ","End":"00:13.140","Text":"Let\u0027s see if the answer is yes or no for convergence in the L2 norm."},{"Start":"00:13.140 ","End":"00:19.380","Text":"We need to check that the functions fn really are in the L2 space."},{"Start":"00:19.380 ","End":"00:24.885","Text":"In other words, that the norm squared is easier to compute"},{"Start":"00:24.885 ","End":"00:31.380","Text":"is finite and the norm squared is the integral of absolute value of fn squared."},{"Start":"00:31.380 ","End":"00:33.135","Text":"Let\u0027s see what that is."},{"Start":"00:33.135 ","End":"00:36.630","Text":"This integral from minus infinity to infinity as before,"},{"Start":"00:36.630 ","End":"00:40.650","Text":"we can just take from n to n plus 1 over n because"},{"Start":"00:40.650 ","End":"00:45.360","Text":"it\u0027s 0 outside and it\u0027s 1 in this interval or 1 squared."},{"Start":"00:45.360 ","End":"00:47.780","Text":"This is equal to, as before,"},{"Start":"00:47.780 ","End":"00:49.670","Text":"n plus 1 over n minus n,"},{"Start":"00:49.670 ","End":"00:52.565","Text":"which is 1 over n, which is finite."},{"Start":"00:52.565 ","End":"00:56.570","Text":"The fn functions do belong to L2."},{"Start":"00:56.570 ","End":"00:58.025","Text":"Now if they converge,"},{"Start":"00:58.025 ","End":"01:01.145","Text":"then they\u0027re going to converge to 0,"},{"Start":"01:01.145 ","End":"01:03.460","Text":"same as the pointwise limit."},{"Start":"01:03.460 ","End":"01:07.335","Text":"Of course, the 0 function belongs to the space."},{"Start":"01:07.335 ","End":"01:09.405","Text":"It is the 0 function after all."},{"Start":"01:09.405 ","End":"01:16.205","Text":"What we have to show is that the limit of the difference of the norms is 0."},{"Start":"01:16.205 ","End":"01:17.690","Text":"Well, instead of the norm,"},{"Start":"01:17.690 ","End":"01:21.245","Text":"we take the norm squared it\u0027s easier to compute."},{"Start":"01:21.245 ","End":"01:28.240","Text":"Let\u0027s see the limit of the difference of the functions norm squared."},{"Start":"01:28.240 ","End":"01:31.245","Text":"We can ignore the minus f because f is 0."},{"Start":"01:31.245 ","End":"01:34.470","Text":"We want the norm squared in L2."},{"Start":"01:34.470 ","End":"01:37.130","Text":"We already computed this over here,"},{"Start":"01:37.130 ","End":"01:43.030","Text":"that this is 1 over n. The limit as n goes to infinity of 1 over n is 0."},{"Start":"01:43.030 ","End":"01:44.490","Text":"We\u0027re okay."},{"Start":"01:44.490 ","End":"01:50.195","Text":"Fn does tend to f tends to 0 in the norm of L2."},{"Start":"01:50.195 ","End":"01:51.905","Text":"The answer is yes,"},{"Start":"01:51.905 ","End":"01:56.550","Text":"and that concludes part d and the whole exercise."}],"ID":28678},{"Watched":false,"Name":"Exercise 4 - Part a","Duration":"2m 3s","ChapterTopicVideoID":27499,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.500","Text":"In this exercise, we have a sequence of functions F_n,"},{"Start":"00:04.500 ","End":"00:12.690","Text":"which is defined on a semi-infinite interval from 1 to infinity as follows."},{"Start":"00:12.690 ","End":"00:15.180","Text":"I won\u0027t read it out."},{"Start":"00:15.180 ","End":"00:22.170","Text":"The function Chi n is defined as 1 if x"},{"Start":"00:22.170 ","End":"00:28.950","Text":"is in this interval from n minus 1 over n^2 to n plus 1 over n^2, and 0 otherwise."},{"Start":"00:28.950 ","End":"00:31.890","Text":"Alpha is some real parameter."},{"Start":"00:31.890 ","End":"00:34.620","Text":"Then we have 3 questions,"},{"Start":"00:34.620 ","End":"00:36.210","Text":"though here we\u0027ll just do part a,"},{"Start":"00:36.210 ","End":"00:40.515","Text":"for which values of Alpha does F_n converge pointwise?"},{"Start":"00:40.515 ","End":"00:43.585","Text":"If it does converge, what is the pointwise limit?"},{"Start":"00:43.585 ","End":"00:48.290","Text":"Similarly, for uniform convergence in part b,"},{"Start":"00:48.290 ","End":"00:51.545","Text":"and convergence in the norm of L^1 for part c,"},{"Start":"00:51.545 ","End":"00:54.775","Text":"but here we\u0027ll just focus on part a."},{"Start":"00:54.775 ","End":"00:58.535","Text":"Let x be in our interval from 1 to infinity."},{"Start":"00:58.535 ","End":"01:01.595","Text":"Choose some natural number N,"},{"Start":"01:01.595 ","End":"01:03.845","Text":"bigger than x plus 1."},{"Start":"01:03.845 ","End":"01:06.740","Text":"If little n is bigger or equal to N,"},{"Start":"01:06.740 ","End":"01:09.575","Text":"then it doesn\u0027t belong to this interval,"},{"Start":"01:09.575 ","End":"01:10.730","Text":"n minus 1 over n^2,"},{"Start":"01:10.730 ","End":"01:12.515","Text":"n plus 1 over n^2,"},{"Start":"01:12.515 ","End":"01:17.650","Text":"and so Chi n(x) is 0."},{"Start":"01:17.650 ","End":"01:20.535","Text":"Let\u0027s go back and look."},{"Start":"01:20.535 ","End":"01:26.030","Text":"Yeah, f_n(x) is equal to just this part,"},{"Start":"01:26.030 ","End":"01:27.500","Text":"because if Chi n is 0,"},{"Start":"01:27.500 ","End":"01:31.460","Text":"this part disappears and 1 minus this is just 1,"},{"Start":"01:31.460 ","End":"01:33.745","Text":"so we\u0027re left with just this."},{"Start":"01:33.745 ","End":"01:39.590","Text":"This is equal to 1 over x plus 1 over n. When n goes to infinity,"},{"Start":"01:39.590 ","End":"01:44.990","Text":"this part goes to 0 and we\u0027re left with 1 over x."},{"Start":"01:44.990 ","End":"01:50.740","Text":"Yes, it converges pointwise to the function f(x) equals 1 over x,"},{"Start":"01:50.740 ","End":"01:53.230","Text":"let\u0027s just write that out."},{"Start":"01:53.230 ","End":"01:56.880","Text":"Note that Alpha didn\u0027t come into the picture anywhere,"},{"Start":"01:56.880 ","End":"02:00.270","Text":"so this is true for all values of Alpha,"},{"Start":"02:00.270 ","End":"02:03.550","Text":"and that answers part a."}],"ID":28679},{"Watched":false,"Name":"Exercise 4 - Part b","Duration":"4m 57s","ChapterTopicVideoID":27500,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.790","Text":"Now part B, where we\u0027re going to check for uniform convergence."},{"Start":"00:05.790 ","End":"00:09.420","Text":"Note that this interval,"},{"Start":"00:09.420 ","End":"00:13.905","Text":"we can write the condition on x slightly differently."},{"Start":"00:13.905 ","End":"00:19.320","Text":"We could say that x minus n in absolute value is less than or equal"},{"Start":"00:19.320 ","End":"00:25.770","Text":"to 1 here and bigger than 1 in the otherwise condition."},{"Start":"00:25.770 ","End":"00:28.740","Text":"We can rewrite fn(x)."},{"Start":"00:28.740 ","End":"00:33.030","Text":"Looking here, if Chi(n) is 1,"},{"Start":"00:33.030 ","End":"00:34.740","Text":"then this is 0,"},{"Start":"00:34.740 ","End":"00:39.540","Text":"so we just get n^ Alpha because this is then 1."},{"Start":"00:39.540 ","End":"00:44.210","Text":"If Chi(n) is 0,"},{"Start":"00:44.210 ","End":"00:46.640","Text":"then we just get this bit."},{"Start":"00:46.640 ","End":"00:50.250","Text":"It\u0027s either this piece or the n^Alpha,"},{"Start":"00:50.290 ","End":"00:55.250","Text":"according to whether this condition holds or it doesn\u0027t hold."},{"Start":"00:55.250 ","End":"00:58.535","Text":"Maybe a graph of fn will help."},{"Start":"00:58.535 ","End":"01:01.670","Text":"Here\u0027s a picture."},{"Start":"01:01.670 ","End":"01:07.340","Text":"In this part, from n minus 1^2(n) plus 1^2."},{"Start":"01:07.340 ","End":"01:09.925","Text":"A value comes from n^Alpha,"},{"Start":"01:09.925 ","End":"01:14.100","Text":"which is a constant as far as x goes."},{"Start":"01:14.100 ","End":"01:17.120","Text":"The other part, x plus 1^-1,"},{"Start":"01:17.120 ","End":"01:20.505","Text":"it\u0027s a bit like 1/x."},{"Start":"01:20.505 ","End":"01:24.289","Text":"This is it, but this holds outside this interval."},{"Start":"01:24.289 ","End":"01:29.340","Text":"Now, suppose that the sequence fn converges uniformly,"},{"Start":"01:29.340 ","End":"01:33.435","Text":"then the limit would be 1/x like in the point-wise limit."},{"Start":"01:33.435 ","End":"01:39.060","Text":"Let\u0027s see if fn(x) does converge to 1/x uniformly,"},{"Start":"01:39.060 ","End":"01:43.475","Text":"or let\u0027s see for which Alpha it does because it depends on Alpha."},{"Start":"01:43.475 ","End":"01:46.220","Text":"Uniform convergence means that the supremum over"},{"Start":"01:46.220 ","End":"01:51.605","Text":"the whole interval of fn minus f goes to 0."},{"Start":"01:51.605 ","End":"01:55.670","Text":"At the supremum, we can break it up into 2 parts."},{"Start":"01:55.670 ","End":"01:59.930","Text":"Because if we\u0027re on the part where absolute value of x"},{"Start":"01:59.930 ","End":"02:05.500","Text":"minus n is less than or equal to 1^2, that\u0027s one thing."},{"Start":"02:05.500 ","End":"02:10.505","Text":"We have a formula for this and outside this."},{"Start":"02:10.505 ","End":"02:12.740","Text":"But with the additional condition,"},{"Start":"02:12.740 ","End":"02:14.630","Text":"of course, that x is bigger or equal to 1."},{"Start":"02:14.630 ","End":"02:17.040","Text":"We have another formula."},{"Start":"02:17.420 ","End":"02:23.585","Text":"Here, like we saw, fn(x) is n^Alpha."},{"Start":"02:23.585 ","End":"02:26.780","Text":"F(x) is 1/x in both cases."},{"Start":"02:26.780 ","End":"02:33.040","Text":"The other part, fn(x) is 1/x plus 1."},{"Start":"02:33.040 ","End":"02:36.480","Text":"I should have said, we take the maximum of these"},{"Start":"02:36.480 ","End":"02:39.470","Text":"2 because the superman could occur here or here,"},{"Start":"02:39.470 ","End":"02:43.810","Text":"and we want the biggest of these 2."},{"Start":"02:43.810 ","End":"02:47.090","Text":"Now, we\u0027ll compute each of the 2 parts separately."},{"Start":"02:47.090 ","End":"02:51.050","Text":"First, we\u0027ll compute the part that was colored in blue,"},{"Start":"02:51.050 ","End":"02:55.315","Text":"and later we\u0027ll do the part that was in red."},{"Start":"02:55.315 ","End":"03:03.945","Text":"This supremum, 1/x plus 1 minus 1/x to a bit of algebra here."},{"Start":"03:03.945 ","End":"03:11.505","Text":"More algebra and we end up with 1(x) plus 1 times x."},{"Start":"03:11.505 ","End":"03:17.375","Text":"This is a decreasing function because the denominator is increasing."},{"Start":"03:17.375 ","End":"03:18.440","Text":"When x gets bigger,"},{"Start":"03:18.440 ","End":"03:20.705","Text":"each of these parts x gets bigger,"},{"Start":"03:20.705 ","End":"03:22.780","Text":"then n(x) plus 1 gets bigger."},{"Start":"03:22.780 ","End":"03:27.125","Text":"Because it\u0027s decreasing the supremum occurs at the very leftmost part,"},{"Start":"03:27.125 ","End":"03:29.315","Text":"which is where x=1."},{"Start":"03:29.315 ","End":"03:32.990","Text":"When x is 1, it\u0027s 1 plus 1."},{"Start":"03:32.990 ","End":"03:36.560","Text":"That\u0027s this part. The other part,"},{"Start":"03:36.560 ","End":"03:45.285","Text":"n^Alpha minus 1/x is the supremum of this here."},{"Start":"03:45.285 ","End":"03:48.705","Text":"This is increasing."},{"Start":"03:48.705 ","End":"03:54.000","Text":"You could just say that n^Alpha is a constant regarding x and 1/x decreases,"},{"Start":"03:54.000 ","End":"03:56.055","Text":"so minus 1/x increases."},{"Start":"03:56.055 ","End":"04:00.875","Text":"Here\u0027s a check with the derivative and because it\u0027s increasing,"},{"Start":"04:00.875 ","End":"04:05.719","Text":"the supremum occurs at the rightmost endpoint,"},{"Start":"04:05.719 ","End":"04:08.795","Text":"which is at n plus 1^2."},{"Start":"04:08.795 ","End":"04:12.580","Text":"This comes out to be the following expression."},{"Start":"04:12.580 ","End":"04:16.370","Text":"Now, we need to take the maximum of this and this."},{"Start":"04:16.370 ","End":"04:19.360","Text":"The supremum is the maximum of these 2."},{"Start":"04:19.360 ","End":"04:22.220","Text":"The question is, does this tend to"},{"Start":"04:22.220 ","End":"04:26.395","Text":"0 or under what conditions on Alpha does this tend to 0?"},{"Start":"04:26.395 ","End":"04:28.605","Text":"This part tends to 0,"},{"Start":"04:28.605 ","End":"04:30.645","Text":"this part tends to 0,"},{"Start":"04:30.645 ","End":"04:34.010","Text":"so the question is under what conditions on Alpha does"},{"Start":"04:34.010 ","End":"04:37.840","Text":"n^Alpha tend to 0 as n goes to infinity?"},{"Start":"04:37.840 ","End":"04:41.000","Text":"The answer is Alpha has to be negative."},{"Start":"04:41.000 ","End":"04:43.445","Text":"If Alpha 0 it\u0027s a constant,"},{"Start":"04:43.445 ","End":"04:45.665","Text":"doesn\u0027t go to 0, and if Alpha is positive,"},{"Start":"04:45.665 ","End":"04:47.330","Text":"this goes to infinity."},{"Start":"04:47.330 ","End":"04:50.620","Text":"Only for Alpha negative does this go to 0."},{"Start":"04:50.620 ","End":"04:53.015","Text":"That\u0027s the answer to part b."},{"Start":"04:53.015 ","End":"04:57.630","Text":"The sequence converges uniformly for negative Alpha."}],"ID":28680},{"Watched":false,"Name":"Exercise 4 - Part c","Duration":"1m 25s","ChapterTopicVideoID":27501,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.170 ","End":"00:03.630","Text":"Now we come to the last part,"},{"Start":"00:03.630 ","End":"00:11.130","Text":"Part C. Which values of Alpha does f_n converge in the norm of L^1 of the interval?"},{"Start":"00:11.130 ","End":"00:16.675","Text":"Turns out that the functions f_n don\u0027t even belong to L^1,"},{"Start":"00:16.675 ","End":"00:20.345","Text":"so you can\u0027t even talk about convergence."},{"Start":"00:20.345 ","End":"00:22.600","Text":"This is regardless of Alpha."},{"Start":"00:22.600 ","End":"00:27.180","Text":"Let\u0027s show this. Here again, is the picture."},{"Start":"00:27.180 ","End":"00:34.595","Text":"The norm in L^1 is the integral of the absolute value of f_n(x)|dx."},{"Start":"00:34.595 ","End":"00:38.180","Text":"Here we don\u0027t need the absolute value because it is non-negative."},{"Start":"00:38.180 ","End":"00:44.569","Text":"Now, this is bigger or equal to the integral on just the last part."},{"Start":"00:44.569 ","End":"00:48.630","Text":"Just from this point onwards."},{"Start":"00:49.270 ","End":"00:52.100","Text":"Here we know what the value is."},{"Start":"00:52.100 ","End":"00:55.355","Text":"The value is 1/x plus 1."},{"Start":"00:55.355 ","End":"00:57.605","Text":"We have this integral."},{"Start":"00:57.605 ","End":"01:04.560","Text":"The integral of 1/x plus a constant is natural log of x plus the constant,"},{"Start":"01:04.560 ","End":"01:08.945","Text":"and we need to evaluate it from n plus 1^2 to infinity."},{"Start":"01:08.945 ","End":"01:10.550","Text":"This thing goes to infinity,"},{"Start":"01:10.550 ","End":"01:14.090","Text":"the natural log also goes to infinity, so it\u0027s infinity."},{"Start":"01:14.090 ","End":"01:17.900","Text":"This norm is not finite, and so, yeah,"},{"Start":"01:17.900 ","End":"01:21.920","Text":"there\u0027s no talk of convergence regardless of Alpha."},{"Start":"01:21.920 ","End":"01:26.310","Text":"That concludes Part C and this exercise."}],"ID":28681},{"Watched":false,"Name":"Exercise 5 - Part a","Duration":"3m 4s","ChapterTopicVideoID":27512,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.860","Text":"This exercise has 3 parts."},{"Start":"00:02.860 ","End":"00:08.505","Text":"Read each part as we come to it and start with Part A."},{"Start":"00:08.505 ","End":"00:12.660","Text":"We have to prove that if f is continuous on the closed interval a,"},{"Start":"00:12.660 ","End":"00:16.155","Text":"b, then the following inequality holds,"},{"Start":"00:16.155 ","End":"00:17.721","Text":"won\u0027t read it out,"},{"Start":"00:17.721 ","End":"00:26.385","Text":"and this is a reminder of what the inner product is and what the norm is in L2, a, b."},{"Start":"00:26.385 ","End":"00:28.845","Text":"We\u0027ll start with Part A."},{"Start":"00:28.845 ","End":"00:32.310","Text":"This is the left-hand side of the inequality in Part A."},{"Start":"00:32.310 ","End":"00:33.870","Text":"First of all, it is less than or equal to."},{"Start":"00:33.870 ","End":"00:36.870","Text":"We can put the absolute value inside the integral."},{"Start":"00:36.870 ","End":"00:38.520","Text":"It\u0027s a standard thing."},{"Start":"00:38.520 ","End":"00:40.125","Text":"Now here\u0027s the trick."},{"Start":"00:40.125 ","End":"00:42.815","Text":"Instead of the integral from a to x,"},{"Start":"00:42.815 ","End":"00:47.915","Text":"I can replace this by an integral from a to b in the following way."},{"Start":"00:47.915 ","End":"00:53.375","Text":"Multiply absolute value of f of t by a function h of t,"},{"Start":"00:53.375 ","End":"00:59.990","Text":"which is 1 between a and x and 0 from x to b."},{"Start":"00:59.990 ","End":"01:04.220","Text":"The reason this works is that the integral from"},{"Start":"01:04.220 ","End":"01:08.660","Text":"a to b is the integral from a to x plus the integral from x to b."},{"Start":"01:08.660 ","End":"01:10.265","Text":"Now, in this part,"},{"Start":"01:10.265 ","End":"01:11.920","Text":"h is equal to 1,"},{"Start":"01:11.920 ","End":"01:16.145","Text":"and in this part, h is equal to 0."},{"Start":"01:16.145 ","End":"01:20.030","Text":"Well, except that the single point x itself,"},{"Start":"01:20.030 ","End":"01:22.180","Text":"but a single point doesn\u0027t matter."},{"Start":"01:22.180 ","End":"01:26.690","Text":"Now this one just drops off, and the 0,"},{"Start":"01:26.690 ","End":"01:28.640","Text":"we eliminate this term altogether and we\u0027re"},{"Start":"01:28.640 ","End":"01:31.010","Text":"left with the integral of just absolute value of"},{"Start":"01:31.010 ","End":"01:37.055","Text":"f. We can put a bar over h of t because h is real."},{"Start":"01:37.055 ","End":"01:41.005","Text":"I\u0027m heading for an inner product that\u0027s why I\u0027m doing that."},{"Start":"01:41.005 ","End":"01:46.730","Text":"This is equal to the inner product of absolute value of f with h bar."},{"Start":"01:46.730 ","End":"01:50.090","Text":"Now, I want to put bars around this absolute value,"},{"Start":"01:50.090 ","End":"01:54.890","Text":"and I can do that because both of these functions are non-negative so"},{"Start":"01:54.890 ","End":"01:56.990","Text":"the inner product is going to be"},{"Start":"01:56.990 ","End":"02:00.610","Text":"non-negative and this integral is going to be non-negative."},{"Start":"02:00.610 ","End":"02:02.450","Text":"Put bars around it."},{"Start":"02:02.450 ","End":"02:06.710","Text":"The reason I\u0027m doing that is because I want to use the Cauchy-Schwarz inequality"},{"Start":"02:06.710 ","End":"02:11.945","Text":"and say that this is less than or equal to norm of this times norm of this."},{"Start":"02:11.945 ","End":"02:17.420","Text":"Also, we can drop the absolute value here and the bar on"},{"Start":"02:17.420 ","End":"02:20.690","Text":"the edge so it\u0027s a norm of f and not norm of"},{"Start":"02:20.690 ","End":"02:24.125","Text":"absolute value of f and the norm of h instead of h bar."},{"Start":"02:24.125 ","End":"02:26.060","Text":"Now, the norm of h,"},{"Start":"02:26.060 ","End":"02:32.660","Text":"we can write as a square root of the integral of h squared on the interval a b,"},{"Start":"02:32.660 ","End":"02:35.770","Text":"where this is what h is a reminder."},{"Start":"02:35.770 ","End":"02:42.735","Text":"This is equal to the norm of f times the integral of 1 from a to x,"},{"Start":"02:42.735 ","End":"02:45.030","Text":"because h is 0 from x to b,"},{"Start":"02:45.030 ","End":"02:46.830","Text":"so we don\u0027t need the path from x to b,"},{"Start":"02:46.830 ","End":"02:54.300","Text":"just from a to x where it\u0027s 1 and this integral is t between the limits a and x,"},{"Start":"02:54.300 ","End":"02:56.240","Text":"so substitute x, substitute a, and subtract."},{"Start":"02:56.240 ","End":"02:58.925","Text":"We get x minus a and there\u0027s a square root here"},{"Start":"02:58.925 ","End":"03:04.410","Text":"and this is what we had to show so that concludes part A."}],"ID":28682},{"Watched":false,"Name":"Exercise 5 - Part b","Duration":"1m 19s","ChapterTopicVideoID":27513,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.710","Text":"Now we come to part b of this exercise."},{"Start":"00:04.710 ","End":"00:10.215","Text":"We have to show that if f_n converges to f in the L^2 norm,"},{"Start":"00:10.215 ","End":"00:13.680","Text":"then it also converges to f in the L^1 norm."},{"Start":"00:13.680 ","End":"00:18.870","Text":"Here\u0027s a reminder of what the L^1 norm is. Let\u0027s solve it."},{"Start":"00:18.870 ","End":"00:27.405","Text":"In other words, we have to show that if the norm of f_n minus f in L^2 tends to 0,"},{"Start":"00:27.405 ","End":"00:33.330","Text":"then the norm of f_n minus f in L^1 also tends to 0."},{"Start":"00:33.330 ","End":"00:36.240","Text":"We\u0027ll use the results of part a,"},{"Start":"00:36.240 ","End":"00:40.815","Text":"and if you replace x by b in part a,"},{"Start":"00:40.815 ","End":"00:42.930","Text":"then we get the following."},{"Start":"00:42.930 ","End":"00:46.120","Text":"Just here was x, now it\u0027s b."},{"Start":"00:46.720 ","End":"00:49.460","Text":"You want to show that this tends to 0."},{"Start":"00:49.460 ","End":"00:50.630","Text":"Well, let\u0027s evaluate it."},{"Start":"00:50.630 ","End":"00:56.480","Text":"This is equal to the integral of f_n(t) minus f(t) dt, absolute value here."},{"Start":"00:56.480 ","End":"00:59.930","Text":"This is less than or equal to using this,"},{"Start":"00:59.930 ","End":"01:04.310","Text":"square root of b minus a times the norm of f_n minus f in L^2."},{"Start":"01:04.310 ","End":"01:07.290","Text":"Now because this tends to 0,"},{"Start":"01:07.290 ","End":"01:08.700","Text":"I mean, that\u0027s given,"},{"Start":"01:08.700 ","End":"01:11.900","Text":"then so does this because this is just a constant,"},{"Start":"01:11.900 ","End":"01:15.290","Text":"so it\u0027s less than or equal to a constant times something going to 0,"},{"Start":"01:15.290 ","End":"01:17.300","Text":"so it also goes to 0."},{"Start":"01:17.300 ","End":"01:20.310","Text":"That concludes part b."}],"ID":28683},{"Watched":false,"Name":"Exercise 5 - Part c","Duration":"5m 17s","ChapterTopicVideoID":27514,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.305","Text":"In Part B, we showed that convergence in L^2 implies convergence in L^1,"},{"Start":"00:07.305 ","End":"00:10.000","Text":"but this applies for a compact interval a, b,"},{"Start":"00:10.000 ","End":"00:13.120","Text":"it doesn\u0027t hold for infinite or open intervals."},{"Start":"00:13.120 ","End":"00:16.435","Text":"Now the other way round, it doesn\u0027t apply."},{"Start":"00:16.435 ","End":"00:20.650","Text":"This convergence is strictly stronger than this convergence and we\u0027ll give"},{"Start":"00:20.650 ","End":"00:25.510","Text":"an example of a sequence that converges in L^1,"},{"Start":"00:25.510 ","End":"00:28.030","Text":"but it doesn\u0027t converge in L^2."},{"Start":"00:28.030 ","End":"00:30.190","Text":"Let\u0027s be specific and take a,"},{"Start":"00:30.190 ","End":"00:36.435","Text":"b as the interval from 0-1 and we\u0027ll take the sequence f_n,"},{"Start":"00:36.435 ","End":"00:38.730","Text":"as in the picture,"},{"Start":"00:38.730 ","End":"00:43.305","Text":"here it is, and f is 0 function."},{"Start":"00:43.305 ","End":"00:45.875","Text":"F_n, if we write it out,"},{"Start":"00:45.875 ","End":"00:51.890","Text":"this has a slope of n over 1/2_n^2,"},{"Start":"00:51.890 ","End":"00:53.165","Text":"which is 2n^3,"},{"Start":"00:53.165 ","End":"00:56.530","Text":"and then multiply it by x because it\u0027s a line through 0,"},{"Start":"00:56.530 ","End":"01:00.350","Text":"and this one we have a slope of minus 2n^3,"},{"Start":"01:00.350 ","End":"01:03.245","Text":"and we also have a point that it passes through."},{"Start":"01:03.245 ","End":"01:08.090","Text":"You can check by plugging in x equals 1 over 2n^2 that we"},{"Start":"01:08.090 ","End":"01:12.710","Text":"really do get n. Then after 1^2,"},{"Start":"01:12.710 ","End":"01:16.700","Text":"the function is 0 and just a technical point,"},{"Start":"01:16.700 ","End":"01:18.800","Text":"I\u0027ve defined this on closed intervals."},{"Start":"01:18.800 ","End":"01:22.910","Text":"They have an overlap at 1 point not the overlap they\u0027re"},{"Start":"01:22.910 ","End":"01:27.450","Text":"both equal so it doesn\u0027t matter for this point and for this point,"},{"Start":"01:27.450 ","End":"01:30.710","Text":"which of the 2 we take on the left or on the right?"},{"Start":"01:30.710 ","End":"01:34.035","Text":"Now, I claim that f_n tends to f,"},{"Start":"01:34.035 ","End":"01:37.210","Text":"which is the 0 function point-wise."},{"Start":"01:37.210 ","End":"01:46.100","Text":"Now here\u0027s an animation just to help the intuition to see why as n goes to infinity,"},{"Start":"01:46.100 ","End":"01:51.350","Text":"the functions f_n tend towards the 0 function."},{"Start":"01:51.350 ","End":"01:53.575","Text":"Well, I hope it helps."},{"Start":"01:53.575 ","End":"01:55.805","Text":"Lets split in 2 cases."},{"Start":"01:55.805 ","End":"01:57.935","Text":"If x is positive,"},{"Start":"01:57.935 ","End":"02:01.459","Text":"then if n is big enough,"},{"Start":"02:01.459 ","End":"02:06.050","Text":"say n is bigger than 1 over square root of x and 1^2 is less than"},{"Start":"02:06.050 ","End":"02:10.500","Text":"x so all this pyramid is all to the left"},{"Start":"02:10.500 ","End":"02:20.190","Text":"of x. F_n of x is 0 for big N so f_n of x tends to 0 and if x is equal to 0,"},{"Start":"02:20.190 ","End":"02:24.750","Text":"then f_n of 0 is 0 and f of 0 is 0 and certainly, that tends also,"},{"Start":"02:24.750 ","End":"02:32.555","Text":"so in general, f_n of x tends to f of x pointwise in the interval from 0-1."},{"Start":"02:32.555 ","End":"02:35.620","Text":"Now I claim that this really does"},{"Start":"02:35.620 ","End":"02:39.580","Text":"converge in the L_1 norm f_n converges to f. In other words that"},{"Start":"02:39.580 ","End":"02:43.690","Text":"this norm of the difference goes to 0 and further I"},{"Start":"02:43.690 ","End":"02:48.085","Text":"will show that it doesn\u0027t converge to 0 in the L_2 norm."},{"Start":"02:48.085 ","End":"02:50.690","Text":"Let\u0027s take care of this part first."},{"Start":"02:50.690 ","End":"02:52.595","Text":"In the L_1 norm,"},{"Start":"02:52.595 ","End":"02:58.010","Text":"this is equal to the integral from 0-1 of the absolute value of the function,"},{"Start":"02:58.010 ","End":"03:00.590","Text":"and this integral we can do by geometry."},{"Start":"03:00.590 ","End":"03:02.704","Text":"It\u0027s the area under the curve."},{"Start":"03:02.704 ","End":"03:06.980","Text":"This part is 0, so we just have the part under the triangle and it\u0027s"},{"Start":"03:06.980 ","End":"03:11.535","Text":"a half base times the height, which is half."},{"Start":"03:11.535 ","End":"03:17.160","Text":"The base is 1^2 and the height is n so it comes out to"},{"Start":"03:17.160 ","End":"03:23.835","Text":"be 1/2n and this does go to 0 as n goes to infinity."},{"Start":"03:23.835 ","End":"03:30.005","Text":"Next, we\u0027ll evaluate the norm of this difference in L_2."},{"Start":"03:30.005 ","End":"03:33.830","Text":"This is equal to the square root of the integral,"},{"Start":"03:33.830 ","End":"03:40.910","Text":"the absolute value squared and f is 0 so we just have this."},{"Start":"03:40.910 ","End":"03:45.500","Text":"I can even drop the absolute value because f_n is always positive."},{"Start":"03:45.500 ","End":"03:48.020","Text":"Since the integral is non-negative,"},{"Start":"03:48.020 ","End":"03:51.500","Text":"if we take the integral not from 0-1,"},{"Start":"03:51.500 ","End":"03:55.520","Text":"but only from 0-1/2n^2 it will only get"},{"Start":"03:55.520 ","End":"03:59.960","Text":"smaller possibly so this is bigger or equal to this."},{"Start":"03:59.960 ","End":"04:03.710","Text":"This is also because it\u0027s easier to compute the integral"},{"Start":"04:03.710 ","End":"04:07.355","Text":"over this spot where we have a single definition for f_n."},{"Start":"04:07.355 ","End":"04:11.575","Text":"In fact, it\u0027s equal to 2n^3x if you go back and look,"},{"Start":"04:11.575 ","End":"04:15.575","Text":"and this is equal to 2n^3 ^2,"},{"Start":"04:15.575 ","End":"04:17.885","Text":"and then the square root is still 2n^3,"},{"Start":"04:17.885 ","End":"04:22.170","Text":"and then we have the integral of x squared from"},{"Start":"04:22.170 ","End":"04:26.415","Text":"0-1/2n^2 square root and this integral is just"},{"Start":"04:26.415 ","End":"04:30.805","Text":"equal 2 or the integral of x squared is a third x cubed and you plugin,"},{"Start":"04:30.805 ","End":"04:36.215","Text":"we get 1/3 times this cubed and here let\u0027s see."},{"Start":"04:36.215 ","End":"04:39.815","Text":"N squared cubed is n to the sixth,"},{"Start":"04:39.815 ","End":"04:43.085","Text":"which comes out of the square root is n^3,"},{"Start":"04:43.085 ","End":"04:46.130","Text":"a half cubed is an 1/8 square root,"},{"Start":"04:46.130 ","End":"04:48.025","Text":"and the square root of a third."},{"Start":"04:48.025 ","End":"04:51.545","Text":"This is root a quarter times root of a half."},{"Start":"04:51.545 ","End":"04:56.990","Text":"The quarter comes out as a half cancels with the 2 third times a half is a 6."},{"Start":"04:56.990 ","End":"05:00.965","Text":"Anyway, it comes out to be always bigger or equal to 1/6,"},{"Start":"05:00.965 ","End":"05:06.590","Text":"which is a positive number and so this difference doesn\u0027t tend to 0."},{"Start":"05:06.590 ","End":"05:11.825","Text":"It\u0027s constantly bigger or equal to square root of 6, it never gets to 0."},{"Start":"05:11.825 ","End":"05:15.770","Text":"This concludes Part C of the exercise,"},{"Start":"05:15.770 ","End":"05:18.480","Text":"and that\u0027s all the exercise."}],"ID":28684},{"Watched":false,"Name":"Exercise 6","Duration":"3m 25s","ChapterTopicVideoID":27515,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.330","Text":"In this exercise, V is a normed space,"},{"Start":"00:03.330 ","End":"00:08.670","Text":"and we have to prove what is called the inverse triangle inequality, this one."},{"Start":"00:08.670 ","End":"00:12.450","Text":"This holds for any u and v in the space"},{"Start":"00:12.450 ","End":"00:19.005","Text":"V. Then we\u0027re going to use part a to prove something we\u0027ve already used."},{"Start":"00:19.005 ","End":"00:22.710","Text":"If we have a convergent sequence f_n,"},{"Start":"00:22.710 ","End":"00:23.835","Text":"which tends to f,"},{"Start":"00:23.835 ","End":"00:26.220","Text":"it could be functions but it doesn\u0027t have to be,"},{"Start":"00:26.220 ","End":"00:28.875","Text":"just could be any elements in a normed space."},{"Start":"00:28.875 ","End":"00:31.575","Text":"If we have convergence in the norm,"},{"Start":"00:31.575 ","End":"00:39.305","Text":"then the norm of each of the members of the sequence converges to the norm of the limit."},{"Start":"00:39.305 ","End":"00:44.701","Text":"Let\u0027s start with part a. I start with this equality,"},{"Start":"00:44.701 ","End":"00:47.360","Text":"because u is equal to u minus v plus v,"},{"Start":"00:47.360 ","End":"00:48.470","Text":"and then we can apply"},{"Start":"00:48.470 ","End":"00:53.365","Text":"the regular triangle inequality and get that it\u0027s less than or equal to this."},{"Start":"00:53.365 ","End":"01:00.530","Text":"Then we can bring the norm of v to the other side and get an inequality like so."},{"Start":"01:00.530 ","End":"01:03.770","Text":"Similarly, if you replace u by v and v by u,"},{"Start":"01:03.770 ","End":"01:06.090","Text":"just switch them,"},{"Start":"01:06.090 ","End":"01:07.360","Text":"you get this,"},{"Start":"01:07.360 ","End":"01:09.362","Text":"which is the same as what\u0027s in here,"},{"Start":"01:09.362 ","End":"01:10.730","Text":"with u and v reversed."},{"Start":"01:10.730 ","End":"01:13.625","Text":"But here we can flip the order,"},{"Start":"01:13.625 ","End":"01:20.585","Text":"norm of v minus u is the same as norm of u minus v. Now, from this inequality,"},{"Start":"01:20.585 ","End":"01:23.800","Text":"we can multiply both sides by minus 1,"},{"Start":"01:23.800 ","End":"01:25.175","Text":"forget the middle bit,"},{"Start":"01:25.175 ","End":"01:27.140","Text":"multiply this by minus 1,"},{"Start":"01:27.140 ","End":"01:29.690","Text":"so we have norm of u minus norm of v,"},{"Start":"01:29.690 ","End":"01:31.850","Text":"and this put a minus in front."},{"Start":"01:31.850 ","End":"01:36.733","Text":"But then we have to switch the direction of the inequality,"},{"Start":"01:36.733 ","End":"01:38.674","Text":"because we multiplied by a negative."},{"Start":"01:38.674 ","End":"01:41.980","Text":"Now we have these 2 inequalities."},{"Start":"01:41.980 ","End":"01:49.025","Text":"What we can do, I claim is write it as 1 double inequality like so."},{"Start":"01:49.025 ","End":"01:53.335","Text":"I color-coded it so you can see which inequality is coming from where."},{"Start":"01:53.335 ","End":"01:59.420","Text":"Now, I\u0027m going to use the result from algebra with real numbers that b is"},{"Start":"01:59.420 ","End":"02:05.996","Text":"between minus a and a is the same thing as absolute value of b is less than a,"},{"Start":"02:05.996 ","End":"02:07.825","Text":"less than or equal to."},{"Start":"02:07.825 ","End":"02:13.202","Text":"Take b as norm of v minus norm of v,"},{"Start":"02:13.202 ","End":"02:18.120","Text":"take a as the norm of u minus v,"},{"Start":"02:20.390 ","End":"02:23.355","Text":"just flip sides,"},{"Start":"02:23.355 ","End":"02:24.785","Text":"so we get this,"},{"Start":"02:24.785 ","End":"02:26.900","Text":"and this is what we had to show."},{"Start":"02:26.900 ","End":"02:29.225","Text":"So that does part a."},{"Start":"02:29.225 ","End":"02:31.545","Text":"Now on to part b."},{"Start":"02:31.545 ","End":"02:37.506","Text":"Suppose f_n tends to f in the normed space."},{"Start":"02:37.506 ","End":"02:43.135","Text":"That means that the norm of f_n minus f tends to 0."},{"Start":"02:43.135 ","End":"02:45.350","Text":"My part a,"},{"Start":"02:45.350 ","End":"02:49.805","Text":"we can write this as bigger or equal to the absolute value of norm of"},{"Start":"02:49.805 ","End":"02:54.945","Text":"f_n minus norm of f. Just flip sides."},{"Start":"02:54.945 ","End":"02:57.080","Text":"This is less than or equal to this,"},{"Start":"02:57.080 ","End":"02:59.405","Text":"and this tends to 0,"},{"Start":"02:59.405 ","End":"03:02.824","Text":"which means that this difference tends to 0."},{"Start":"03:02.824 ","End":"03:04.670","Text":"This is a sequence, this is fixed."},{"Start":"03:04.670 ","End":"03:08.000","Text":"If this minus this tends to 0,"},{"Start":"03:08.000 ","End":"03:13.165","Text":"then this sequence tends to this number."},{"Start":"03:13.165 ","End":"03:17.300","Text":"This is what we have or we can rewrite it as just another way,"},{"Start":"03:17.300 ","End":"03:21.440","Text":"if you like, with limit instead of the big arrow."},{"Start":"03:21.440 ","End":"03:24.080","Text":"That proves part b."},{"Start":"03:24.080 ","End":"03:26.550","Text":"So we\u0027re done."}],"ID":28685},{"Watched":false,"Name":"Exercise 7 - Part a","Duration":"2m 31s","ChapterTopicVideoID":27496,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.820","Text":"In this exercise, v is a vector space of functions from the interval a,"},{"Start":"00:05.820 ","End":"00:07.740","Text":"b to the reals,"},{"Start":"00:07.740 ","End":"00:13.545","Text":"such that f is differentiable except possibly at a finite number of points,"},{"Start":"00:13.545 ","End":"00:17.520","Text":"and its derivative is piecewise continuous."},{"Start":"00:17.520 ","End":"00:22.025","Text":"We define an inner product on this space,"},{"Start":"00:22.025 ","End":"00:24.980","Text":"f product with g is f(a),"},{"Start":"00:24.980 ","End":"00:30.400","Text":"g(a) plus the integral from a to b of f\u0027 times g\u0027."},{"Start":"00:30.400 ","End":"00:34.545","Text":"This is known to be an inner product I won\u0027t prove it here"},{"Start":"00:34.545 ","End":"00:41.135","Text":"and let this norm be the derived norm from the inner product."},{"Start":"00:41.135 ","End":"00:46.310","Text":"In this step, we\u0027ll just do Part a we\u0027ll read each part as we come to it."},{"Start":"00:46.310 ","End":"00:50.450","Text":"In Part a, if x naught is in the interval,"},{"Start":"00:50.450 ","End":"00:55.055","Text":"we define a function g_x_naught as follows."},{"Start":"00:55.055 ","End":"00:59.870","Text":"It\u0027s a piecewise function defined as this."},{"Start":"00:59.870 ","End":"01:01.520","Text":"There\u0027s a picture below,"},{"Start":"01:01.520 ","End":"01:05.895","Text":"we\u0027ll get to it and we have to prove that"},{"Start":"01:05.895 ","End":"01:11.780","Text":"these g_x_naught have the property that the inner product of f with g_x_naught,"},{"Start":"01:11.780 ","End":"01:14.845","Text":"gives us exactly f(x)_naught."},{"Start":"01:14.845 ","End":"01:18.749","Text":"I\u0027ll show you a picture of g_x_naught."},{"Start":"01:18.749 ","End":"01:24.140","Text":"It starts up with slope of 1 and then flattens out."},{"Start":"01:24.140 ","End":"01:32.525","Text":"Note that its derivative is 1 up to the point x_naught and 0 from the point x_naught."},{"Start":"01:32.525 ","End":"01:35.690","Text":"Actually, the derivative doesn\u0027t exist"},{"Start":"01:35.690 ","End":"01:40.025","Text":"exactly at x_naught because I fixed that to be precise."},{"Start":"01:40.025 ","End":"01:44.480","Text":"Now we want to compute the inner product of f with g_x_naught."},{"Start":"01:44.480 ","End":"01:50.525","Text":"By the definition, it\u0027s f(a) g_x_naught of a plus the integral this,"},{"Start":"01:50.525 ","End":"01:55.145","Text":"and this is equal to g_x_naught of a is 1."},{"Start":"01:55.145 ","End":"01:57.005","Text":"You can go back and look at the definition on"},{"Start":"01:57.005 ","End":"02:01.850","Text":"the graph and we break this integral up into 2 parts,"},{"Start":"02:01.850 ","End":"02:05.495","Text":"from a to x_naught and from x_naught to b."},{"Start":"02:05.495 ","End":"02:09.540","Text":"This is because g_x_naught is piecewise defined."},{"Start":"02:09.650 ","End":"02:14.115","Text":"On this piece, it\u0027s equal to 1,"},{"Start":"02:14.115 ","End":"02:16.880","Text":"the g\u0027_x_naught from here,"},{"Start":"02:16.880 ","End":"02:19.355","Text":"and here it\u0027s 0."},{"Start":"02:19.355 ","End":"02:23.800","Text":"Now this integral is f(x)_naught minus f(a),"},{"Start":"02:23.800 ","End":"02:28.165","Text":"so this just comes out to be f(x)_naught."},{"Start":"02:28.165 ","End":"02:31.770","Text":"That\u0027s all we had to prove in part a."}],"ID":28686},{"Watched":false,"Name":"Exercise 7 - Part b","Duration":"1m 35s","ChapterTopicVideoID":27497,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.200","Text":"Now we come to part b of this exercise."},{"Start":"00:04.200 ","End":"00:07.710","Text":"We\u0027ll use part a to prove part b."},{"Start":"00:07.710 ","End":"00:11.910","Text":"We have to prove that the absolute value of f(x_naught) is"},{"Start":"00:11.910 ","End":"00:16.335","Text":"less than or equal to square root of b minus a plus 1 times the norm of"},{"Start":"00:16.335 ","End":"00:21.232","Text":"f. The first thing is we use the result of part"},{"Start":"00:21.232 ","End":"00:26.720","Text":"a to write f(x_naught) as the inner product of f with g_x_naught,"},{"Start":"00:26.720 ","End":"00:31.930","Text":"and then the absolute value of both sides."},{"Start":"00:31.930 ","End":"00:35.870","Text":"Next, we\u0027ll use the Cauchy Schwarz inequality"},{"Start":"00:35.870 ","End":"00:41.360","Text":"to estimate this as less than or equal to norm of f, norm of g_x_naught."},{"Start":"00:41.360 ","End":"00:46.285","Text":"The norm of g_x_naught using the inner product,"},{"Start":"00:46.285 ","End":"00:48.830","Text":"this is the induced norm,"},{"Start":"00:48.830 ","End":"00:52.490","Text":"so it\u0027s square root of g_x_naught inner product with itself,"},{"Start":"00:52.490 ","End":"00:56.120","Text":"and then using the definition of the inner product,"},{"Start":"00:56.120 ","End":"00:59.306","Text":"we get this expression."},{"Start":"00:59.306 ","End":"01:04.470","Text":"Recall that g_x_naught derivative is 1 or 0"},{"Start":"01:04.470 ","End":"01:10.625","Text":"accordingly so we can break this integral up into 2 pieces,"},{"Start":"01:10.625 ","End":"01:13.040","Text":"a to x_naught and x_naught to b."},{"Start":"01:13.040 ","End":"01:15.374","Text":"But the second part is just 0,"},{"Start":"01:15.374 ","End":"01:21.920","Text":"so what we get here is 1 plus x_naught minus a."},{"Start":"01:21.920 ","End":"01:25.205","Text":"Since x_naught is less than or equal to b,"},{"Start":"01:25.205 ","End":"01:27.842","Text":"and the square root is an increasing function,"},{"Start":"01:27.842 ","End":"01:32.772","Text":"so we can replace this with 1 plus b minus a."},{"Start":"01:32.772 ","End":"01:35.950","Text":"That concludes part b."}],"ID":28687},{"Watched":false,"Name":"Exercise 7 - Part c","Duration":"1m 45s","ChapterTopicVideoID":27498,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.530","Text":"Now we come to part c which,"},{"Start":"00:03.530 ","End":"00:04.800","Text":"to paraphrase it,"},{"Start":"00:04.800 ","End":"00:12.510","Text":"says that convergence in the norm of V is stronger than uniform convergence."},{"Start":"00:12.510 ","End":"00:16.695","Text":"If f_n converges to f in the norm of V,"},{"Start":"00:16.695 ","End":"00:20.340","Text":"then f_n converges to f uniformly."},{"Start":"00:20.340 ","End":"00:22.080","Text":"Let\u0027s prove that."},{"Start":"00:22.080 ","End":"00:25.470","Text":"We\u0027ll be using the result of part b."},{"Start":"00:25.470 ","End":"00:29.648","Text":"Suppose that f_n minus f tends to 0,"},{"Start":"00:29.648 ","End":"00:31.103","Text":"in the norm,"},{"Start":"00:31.103 ","End":"00:34.240","Text":"that\u0027s what we mean by converging the norm of V."},{"Start":"00:34.240 ","End":"00:37.590","Text":"We have to show that it converges uniformly which means"},{"Start":"00:37.590 ","End":"00:40.470","Text":"that the supremum of the absolute value of"},{"Start":"00:40.470 ","End":"00:44.690","Text":"the difference converges to 0 as n goes to infinity."},{"Start":"00:44.690 ","End":"00:47.420","Text":"Now we\u0027ll, for a particular n,"},{"Start":"00:47.420 ","End":"00:51.260","Text":"let g=f_n minus f. From part b,"},{"Start":"00:51.260 ","End":"00:55.490","Text":"we have that the absolute value of g of x_0 is less than or equal to"},{"Start":"00:55.490 ","End":"01:01.325","Text":"this constant times the norm of g. That means that the absolute value of"},{"Start":"01:01.325 ","End":"01:07.400","Text":"f_n minus f at x_0 is less than or equal to this constant times"},{"Start":"01:07.400 ","End":"01:14.550","Text":"the norm of f_n minus g. This is true for all x_0 in the interval,"},{"Start":"01:14.550 ","End":"01:16.780","Text":"x_0 is any point in that interval."},{"Start":"01:16.780 ","End":"01:22.040","Text":"Just replace x_0 by x so we have the same thing just with x."},{"Start":"01:22.040 ","End":"01:26.135","Text":"That shows that the supremum,"},{"Start":"01:26.135 ","End":"01:27.905","Text":"because it\u0027s true for each x,"},{"Start":"01:27.905 ","End":"01:32.450","Text":"then the supremum is also less than or equal to this."},{"Start":"01:32.450 ","End":"01:40.945","Text":"This is a constant times something which tends to 0 so this also tends to 0."},{"Start":"01:40.945 ","End":"01:46.150","Text":"That concludes part c and we\u0027re done."}],"ID":28688},{"Watched":false,"Name":"Exercise 8 - Part a","Duration":"3m 22s","ChapterTopicVideoID":27502,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.315","Text":"In this exercise, we have a sequence of functions defined on all of the reals."},{"Start":"00:06.315 ","End":"00:11.280","Text":"I\u0027d like to start with a remark about this notation for the Chi function."},{"Start":"00:11.280 ","End":"00:15.435","Text":"This is often used for a characteristic function."},{"Start":"00:15.435 ","End":"00:21.615","Text":"In general if we have any subset of the reals,"},{"Start":"00:21.615 ","End":"00:32.250","Text":"then we define chi sub S of x to be 1 if x belongs to S and 0 otherwise."},{"Start":"00:32.250 ","End":"00:36.525","Text":"That\u0027s useful thing to have often."},{"Start":"00:36.525 ","End":"00:39.710","Text":"Characteristic function, basically it says if it\u0027s in"},{"Start":"00:39.710 ","End":"00:43.945","Text":"the subset or not in the subset 1 if it is 0 otherwise."},{"Start":"00:43.945 ","End":"00:49.465","Text":"In particular, we can apply it to an interval chi sub interval a,"},{"Start":"00:49.465 ","End":"00:57.140","Text":"b and what that does to x is return 1 if x is in the interval and 0 otherwise."},{"Start":"00:57.140 ","End":"01:02.855","Text":"Our sequence f_n is defined in terms of this function as shown."},{"Start":"01:02.855 ","End":"01:03.980","Text":"In part A,"},{"Start":"01:03.980 ","End":"01:12.185","Text":"we have to decide whether the sequence f_n converges in the norm of L1 and in part B,"},{"Start":"01:12.185 ","End":"01:15.310","Text":"same thing, but for the norm of L2,."},{"Start":"01:15.310 ","End":"01:17.210","Text":"Because it\u0027s an infinite interval,"},{"Start":"01:17.210 ","End":"01:20.970","Text":"we can\u0027t conclude that if it converges in a,"},{"Start":"01:20.970 ","End":"01:24.115","Text":"then it converges in b or vice versa."},{"Start":"01:24.115 ","End":"01:26.845","Text":"We just don\u0027t know, we have to check each 1."},{"Start":"01:26.845 ","End":"01:29.485","Text":"We\u0027ll start with the first 1."},{"Start":"01:29.485 ","End":"01:32.980","Text":"The first thing we do is check if these functions even"},{"Start":"01:32.980 ","End":"01:36.670","Text":"belong to L1 of this interval in other words,"},{"Start":"01:36.670 ","End":"01:39.680","Text":"if the norm is less than infinity."},{"Start":"01:39.680 ","End":"01:44.360","Text":"Norm is the integral of the absolute value."},{"Start":"01:46.170 ","End":"01:51.205","Text":"We can drop the absolute value because"},{"Start":"01:51.205 ","End":"01:56.200","Text":"each of these terms is non-negative and also because it\u0027s a finite sum,"},{"Start":"01:56.200 ","End":"02:00.370","Text":"we can switch the order of summation and integration."},{"Start":"02:00.370 ","End":"02:09.140","Text":"Get the sum from 1 to n this integral also take 1 over k^2 in front of the integral."},{"Start":"02:09.140 ","End":"02:15.830","Text":"This chi function is 0 outside the interval 2 to the k,"},{"Start":"02:15.830 ","End":"02:18.440","Text":"2 to the k plus k, and here it\u0027s 1."},{"Start":"02:18.440 ","End":"02:20.315","Text":"This is what we get."},{"Start":"02:20.315 ","End":"02:26.280","Text":"This integral comes out to be 2 to the k plus k minus 2 to the k,"},{"Start":"02:26.280 ","End":"02:33.425","Text":"which is just k. We have harmonic series but finite sum from 1 to n,"},{"Start":"02:33.425 ","End":"02:35.480","Text":"and this is some finite number."},{"Start":"02:35.480 ","End":"02:39.740","Text":"The function f_n for each n belongs to L1."},{"Start":"02:39.740 ","End":"02:43.340","Text":"Now the question is, does it converge?"},{"Start":"02:43.400 ","End":"02:48.105","Text":"Note that the norm of f_n,"},{"Start":"02:48.105 ","End":"02:51.440","Text":"which is the sum from k equals 1 to n,"},{"Start":"02:51.440 ","End":"02:57.665","Text":"tends to infinity because it\u0027s the partial series of the famous harmonic series."},{"Start":"02:57.665 ","End":"02:59.945","Text":"If we write infinity here,"},{"Start":"02:59.945 ","End":"03:03.310","Text":"then the sum is infinity."},{"Start":"03:03.310 ","End":"03:08.090","Text":"The sequence does not converge because we know that if a sequence converges,"},{"Start":"03:08.090 ","End":"03:12.320","Text":"then the sequence of norms also converges and here it doesn\u0027t."},{"Start":"03:12.320 ","End":"03:15.660","Text":"I just wrote this out."},{"Start":"03:15.710 ","End":"03:19.700","Text":"This solves part A where the answer is no,"},{"Start":"03:19.700 ","End":"03:22.860","Text":"the sequence does not converge."}],"ID":28689},{"Watched":false,"Name":"Exercise 8 - Part b1","Duration":"2m 57s","ChapterTopicVideoID":27503,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.300","Text":"Now we come to Part B of the question and I warn you it\u0027s quite lengthy."},{"Start":"00:06.300 ","End":"00:07.919","Text":"Let\u0027s get started."},{"Start":"00:07.919 ","End":"00:13.790","Text":"The first thing to do is to check that the functions f_n even belong to the space L^2."},{"Start":"00:13.790 ","End":"00:17.915","Text":"In other words that the norm in L^2 is finite."},{"Start":"00:17.915 ","End":"00:20.690","Text":"When we check this, we usually check it squared,"},{"Start":"00:20.690 ","End":"00:23.340","Text":"not bother with the square root."},{"Start":"00:23.660 ","End":"00:30.020","Text":"The norm squared of f_n in the L^2 space is"},{"Start":"00:30.020 ","End":"00:36.335","Text":"equal to the integral of absolute value of f_n squared."},{"Start":"00:36.335 ","End":"00:40.580","Text":"That\u0027s equal to the integral of absolute value."},{"Start":"00:40.580 ","End":"00:43.315","Text":"This is f_n squared."},{"Start":"00:43.315 ","End":"00:47.420","Text":"Squared means that it\u0027s multiplied by itself."},{"Start":"00:47.420 ","End":"00:50.630","Text":"This times itself within the integral sign."},{"Start":"00:50.630 ","End":"00:54.800","Text":"We need to use a different index for this one and this one because when they\u0027re"},{"Start":"00:54.800 ","End":"00:59.065","Text":"going to multiply it\u0027s each k here with each j here."},{"Start":"00:59.065 ","End":"01:07.420","Text":"Now, here\u0027s a picture of two of these chi functions for different k and j."},{"Start":"01:07.420 ","End":"01:09.580","Text":"If they\u0027re the same, they exactly overlap."},{"Start":"01:09.580 ","End":"01:10.810","Text":"But if k is different from j,"},{"Start":"01:10.810 ","End":"01:12.835","Text":"there is no overlap at all."},{"Start":"01:12.835 ","End":"01:15.805","Text":"If we multiply this function,"},{"Start":"01:15.805 ","End":"01:22.230","Text":"chi with the k and chi with the 2^j the product is 0."},{"Start":"01:22.230 ","End":"01:24.990","Text":"Where this is 1, this is 0 and where this is 1,"},{"Start":"01:24.990 ","End":"01:27.570","Text":"this is 0, so the product is 0."},{"Start":"01:27.570 ","End":"01:30.295","Text":"That will make the multiplying simpler."},{"Start":"01:30.295 ","End":"01:34.510","Text":"Here I just rearrange the sums and bring them"},{"Start":"01:34.510 ","End":"01:39.580","Text":"outside the integral together with the 1 over k^2 and 1 over j^2."},{"Start":"01:39.580 ","End":"01:47.570","Text":"We really just have the sum of the integral of chi with a k and chi with a j here."},{"Start":"01:47.570 ","End":"01:49.970","Text":"Now we can do what we said,"},{"Start":"01:49.970 ","End":"01:53.420","Text":"that this product is 0 if k is not equal to j,"},{"Start":"01:53.420 ","End":"02:02.115","Text":"so we just get this thing squared and we let j equals k. Simplify that 1 over k^4,"},{"Start":"02:02.115 ","End":"02:05.630","Text":"and then here, chi-squared is either 1 or 0,"},{"Start":"02:05.630 ","End":"02:12.090","Text":"depending on whether we\u0027re in the integral from 2^k to 2^k plus k and there it\u0027s 1."},{"Start":"02:12.320 ","End":"02:16.890","Text":"This integral is just 2^k plus k minus 2^k,"},{"Start":"02:16.890 ","End":"02:22.545","Text":"which is k. k over k^4 is 1 over k^3."},{"Start":"02:22.545 ","End":"02:24.765","Text":"This is some number which is finite."},{"Start":"02:24.765 ","End":"02:28.555","Text":"The f_n are all in L^2."},{"Start":"02:28.555 ","End":"02:35.795","Text":"The pointwise limit of this finite series is just the infinite series,"},{"Start":"02:35.795 ","End":"02:38.315","Text":"change the n to infinity."},{"Start":"02:38.315 ","End":"02:43.535","Text":"This is going to be our candidate for the limit in L^2."},{"Start":"02:43.535 ","End":"02:49.205","Text":"We\u0027re going to show that this f is in L^2 of the real line,"},{"Start":"02:49.205 ","End":"02:54.050","Text":"and that f_n turns to f in L^2 of the line."},{"Start":"02:54.050 ","End":"02:57.900","Text":"Let\u0027s take a break and continue after."}],"ID":28690},{"Watched":false,"Name":"Exercise 8 - Part b2","Duration":"4m 8s","ChapterTopicVideoID":27504,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:02.745","Text":"Back from the break."},{"Start":"00:02.745 ","End":"00:05.910","Text":"Now we\u0027re going to show that f_n converges to f. Just"},{"Start":"00:05.910 ","End":"00:09.735","Text":"note again that f_n is a finite sum from 1 to n,"},{"Start":"00:09.735 ","End":"00:13.615","Text":"whereas f is the infinite sum from 1 to infinity,"},{"Start":"00:13.615 ","End":"00:16.740","Text":"so that the difference f_n minus f or"},{"Start":"00:16.740 ","End":"00:20.550","Text":"the other way around will be the sum from n plus 1 to infinity."},{"Start":"00:20.550 ","End":"00:23.880","Text":"In summary, f_n is the sum from 1 to n,"},{"Start":"00:23.880 ","End":"00:26.595","Text":"f is the sum from 1 to infinity."},{"Start":"00:26.595 ","End":"00:30.630","Text":"We need to show that f_n converges to f by showing"},{"Start":"00:30.630 ","End":"00:35.280","Text":"that f_n minus f in the norm goes to 0."},{"Start":"00:35.280 ","End":"00:40.150","Text":"We take it\u0027s squared so as not to bother with the square root."},{"Start":"00:40.150 ","End":"00:46.805","Text":"F minus f_n is the tail of the series just from n plus 1 to infinity,"},{"Start":"00:46.805 ","End":"00:54.520","Text":"and the norm squared is the integral of the absolute value of the difference squared,"},{"Start":"00:54.520 ","End":"00:57.605","Text":"and this is equal to, like we said,"},{"Start":"00:57.605 ","End":"01:01.535","Text":"the sum from n plus 1 to infinity here."},{"Start":"01:01.535 ","End":"01:05.840","Text":"The sum squared is the sum times itself."},{"Start":"01:05.840 ","End":"01:09.830","Text":"When we do that we change the variable for the index."},{"Start":"01:09.830 ","End":"01:12.410","Text":"Here it\u0027s k, we\u0027ll choose j here,"},{"Start":"01:12.410 ","End":"01:17.980","Text":"so that we can really get 1 from here and 1 from here each time in the double sum."},{"Start":"01:17.980 ","End":"01:19.845","Text":"This is the double sum,"},{"Start":"01:19.845 ","End":"01:21.810","Text":"1 over k^2, 1 over j^2,"},{"Start":"01:21.810 ","End":"01:25.595","Text":"Chi with the k here and Chi with the j here."},{"Start":"01:25.595 ","End":"01:27.395","Text":"But as we saw before,"},{"Start":"01:27.395 ","End":"01:29.390","Text":"if k is not equal to j,"},{"Start":"01:29.390 ","End":"01:32.160","Text":"this gives us the 0 function,"},{"Start":"01:32.160 ","End":"01:36.605","Text":"so we can leave those out and just take when k=j."},{"Start":"01:36.605 ","End":"01:38.615","Text":"We get 1 over k^4,"},{"Start":"01:38.615 ","End":"01:40.670","Text":"and here Chi squared."},{"Start":"01:40.670 ","End":"01:45.605","Text":"Now the idea is to change the summation with the integration."},{"Start":"01:45.605 ","End":"01:48.230","Text":"But because it\u0027s an infinite series,"},{"Start":"01:48.230 ","End":"01:51.364","Text":"we have to show that there is uniform convergence,"},{"Start":"01:51.364 ","End":"01:54.905","Text":"and for that we\u0027re going to use the Weierstrass M-test."},{"Start":"01:54.905 ","End":"02:00.560","Text":"Note that the absolute value of the term here is"},{"Start":"02:00.560 ","End":"02:06.590","Text":"less than or equal to 1 over k^4 because Chi squared is always less than or equal to 1,"},{"Start":"02:06.590 ","End":"02:12.305","Text":"it\u0027s either 0 or 1, and this is a convergent numerical series."},{"Start":"02:12.305 ","End":"02:14.660","Text":"By the Weierstrass M- test,"},{"Start":"02:14.660 ","End":"02:19.130","Text":"the series of functions converges uniformly."},{"Start":"02:19.130 ","End":"02:23.840","Text":"We can switch the order of the integration and the summation and"},{"Start":"02:23.840 ","End":"02:28.370","Text":"get that this is equal to first the sum and then the integral,"},{"Start":"02:28.370 ","End":"02:32.135","Text":"and of course we can pull 1 over k^4 in front of the integral."},{"Start":"02:32.135 ","End":"02:34.040","Text":"Now, as we\u0027ve seen before,"},{"Start":"02:34.040 ","End":"02:38.855","Text":"this integral gives us just k. What we have,"},{"Start":"02:38.855 ","End":"02:45.470","Text":"the k over k^4 becomes 1 over k^3 and we get the sum from n plus 1 to infinity,"},{"Start":"02:45.470 ","End":"02:52.575","Text":"1 over k^3 because the sum from 1 to infinity of 1 over k^3 is convergent,"},{"Start":"02:52.575 ","End":"02:59.080","Text":"the tail of a convergence series goes to 0."},{"Start":"02:59.080 ","End":"03:01.250","Text":"There\u0027s one thing we still haven\u0027t done,"},{"Start":"03:01.250 ","End":"03:06.960","Text":"we haven\u0027t shown that our limit actually does belong to the space L2."},{"Start":"03:06.960 ","End":"03:11.120","Text":"In other words, that the norm or norm squared is less than infinity."},{"Start":"03:11.120 ","End":"03:12.980","Text":"We\u0027re repeating a lot of the work we\u0027ve done already."},{"Start":"03:12.980 ","End":"03:14.765","Text":"Let\u0027s go through this quickly."},{"Start":"03:14.765 ","End":"03:22.805","Text":"This integral is equal to the integral of the sum from 1 to infinity squared,"},{"Start":"03:22.805 ","End":"03:25.790","Text":"and then we write it as the sum times itself,"},{"Start":"03:25.790 ","End":"03:28.840","Text":"but we use a different letter for the index."},{"Start":"03:28.840 ","End":"03:33.105","Text":"Then we get a double sum with k and j."},{"Start":"03:33.105 ","End":"03:35.270","Text":"But when k is not equal to j,"},{"Start":"03:35.270 ","End":"03:36.890","Text":"this product is 0."},{"Start":"03:36.890 ","End":"03:40.650","Text":"We just take a single sum where j=k,"},{"Start":"03:40.650 ","End":"03:43.350","Text":"so we get k^4 and then Chi squared."},{"Start":"03:43.350 ","End":"03:46.520","Text":"Then like before, this converges uniformly"},{"Start":"03:46.520 ","End":"03:50.140","Text":"so we can exchange the integral with the summation."},{"Start":"03:50.140 ","End":"03:52.680","Text":"This integral is k,"},{"Start":"03:52.680 ","End":"04:00.170","Text":"and so we get the sum of 1 over k^3 and this is a convergent series,"},{"Start":"04:00.170 ","End":"04:04.790","Text":"so this sum is less than infinity and that\u0027s what we had to show."},{"Start":"04:04.790 ","End":"04:09.600","Text":"That finally completes this exercise."}],"ID":28691},{"Watched":false,"Name":"Exercise 9 - Part a","Duration":"3m 46s","ChapterTopicVideoID":27505,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.700","Text":"In this exercise, we have a sequence f_n of functions which"},{"Start":"00:05.700 ","End":"00:11.145","Text":"are defined in this space L_2 on the interval from minus Pi to Pi."},{"Start":"00:11.145 ","End":"00:15.390","Text":"This is the definition f_n(x) is a finite sum."},{"Start":"00:15.390 ","End":"00:20.280","Text":"K goes from 1 to n of 1 over root k sine(kx)."},{"Start":"00:20.280 ","End":"00:24.060","Text":"There are 4 parts. We will read each part as we come to it."},{"Start":"00:24.060 ","End":"00:30.450","Text":"Part a, does the sequence f_n converge in the norm of L_2 on this interval?"},{"Start":"00:30.450 ","End":"00:32.265","Text":"The answer is no,"},{"Start":"00:32.265 ","End":"00:35.190","Text":"because the sequence of norms tends to infinity."},{"Start":"00:35.190 ","End":"00:36.570","Text":"That\u0027s one of the basic tests."},{"Start":"00:36.570 ","End":"00:39.720","Text":"If you even suspect that it might not converge,"},{"Start":"00:39.720 ","End":"00:44.240","Text":"you can just check the norms and see that they don\u0027t converge."},{"Start":"00:44.240 ","End":"00:51.995","Text":"Norm of f_n squared is the integral of this sum squared dx,"},{"Start":"00:51.995 ","End":"00:53.360","Text":"and this is equal to,"},{"Start":"00:53.360 ","End":"00:55.895","Text":"to rep the absolute value squared we can"},{"Start":"00:55.895 ","End":"00:58.640","Text":"drop the absolute value and just multiply it by itself."},{"Start":"00:58.640 ","End":"01:02.585","Text":"But remember we changed the index when we have sum times a sum."},{"Start":"01:02.585 ","End":"01:06.235","Text":"This is equal to the double sum on k and j,"},{"Start":"01:06.235 ","End":"01:10.775","Text":"1 over root j 1 of whenever root k, sine(jx)sine(kx)."},{"Start":"01:10.775 ","End":"01:19.700","Text":"Now, it\u0027s a well-known fact that sine(jx) and sine(kx) are orthogonal to each other,"},{"Start":"01:19.700 ","End":"01:23.490","Text":"meaning that the product in L_2,"},{"Start":"01:23.490 ","End":"01:26.785","Text":"which is the integral of 1 times the other,"},{"Start":"01:26.785 ","End":"01:31.450","Text":"is 0 if j doesn\u0027t equal k,"},{"Start":"01:31.450 ","End":"01:34.480","Text":"and Pi if j equals k. There\u0027s actually"},{"Start":"01:34.480 ","End":"01:38.245","Text":"an assumption here that k and j are positive integers,"},{"Start":"01:38.245 ","End":"01:43.228","Text":"which they are because we\u0027re going from 1 to n. I\u0027ll show you a proof at the end."},{"Start":"01:43.228 ","End":"01:46.060","Text":"Though this is equal too."},{"Start":"01:46.060 ","End":"01:50.820","Text":"We just have to take k=j and then from here,"},{"Start":"01:50.820 ","End":"01:52.770","Text":"we put Pi in."},{"Start":"01:52.770 ","End":"01:55.840","Text":"We have this sum and this is equal to,"},{"Start":"01:55.840 ","End":"02:02.500","Text":"pull the Pi out in front and we have 1/k and the sum of the harmonic series,"},{"Start":"02:02.500 ","End":"02:06.170","Text":"the finite sum goes to"},{"Start":"02:06.170 ","End":"02:08.540","Text":"infinity because the sum from 1 to infinity of"},{"Start":"02:08.540 ","End":"02:11.060","Text":"the harmonic series is known to be divergent,"},{"Start":"02:11.060 ","End":"02:12.605","Text":"it goes to infinity."},{"Start":"02:12.605 ","End":"02:21.560","Text":"Now, I owe you a proof that the integral of sine times sine for j not equal to k is 0,"},{"Start":"02:21.560 ","End":"02:22.850","Text":"and at t=Pi,"},{"Start":"02:22.850 ","End":"02:26.045","Text":"if j does equal to k. First of all,"},{"Start":"02:26.045 ","End":"02:29.300","Text":"I\u0027m going to change it from minus Pi to Pi,"},{"Start":"02:29.300 ","End":"02:32.850","Text":"change that to 0 to 2Pi because it\u0027s periodic."},{"Start":"02:32.850 ","End":"02:36.755","Text":"Any interval of length 2Pi will do."},{"Start":"02:36.755 ","End":"02:41.570","Text":"I use this because I found proof for it on the internet and I\u0027ve copied pasted,"},{"Start":"02:41.570 ","End":"02:44.945","Text":"so forgive the different font."},{"Start":"02:44.945 ","End":"02:53.075","Text":"One of the trigonometric identities is sine times sine in terms of cosine is this."},{"Start":"02:53.075 ","End":"02:56.360","Text":"Then when we have sine(nx) sine (mx) we use"},{"Start":"02:56.360 ","End":"03:00.040","Text":"this but with a minus 1/2 and these 2 reversed."},{"Start":"03:00.040 ","End":"03:02.715","Text":"I guess m and n are also flipped."},{"Start":"03:02.715 ","End":"03:05.655","Text":"We get this. You can follow this,"},{"Start":"03:05.655 ","End":"03:08.940","Text":"nothing deep here, just technical."},{"Start":"03:08.940 ","End":"03:12.030","Text":"That does the part for 0."},{"Start":"03:12.030 ","End":"03:15.800","Text":"Then the part for sine times sine is sine squared,"},{"Start":"03:15.800 ","End":"03:18.485","Text":"sine squared nx gives us,"},{"Start":"03:18.485 ","End":"03:21.020","Text":"you can follow this is Pi."},{"Start":"03:21.020 ","End":"03:26.650","Text":"It uses the trigonometric identity that cosine of 2x is 1 minus 2 sine squared x."},{"Start":"03:26.650 ","End":"03:30.110","Text":"This part from here to here is a substitution."},{"Start":"03:30.110 ","End":"03:33.695","Text":"You substitute t=nx,"},{"Start":"03:33.695 ","End":"03:38.950","Text":"but then you switch t back to x and you get this."},{"Start":"03:38.950 ","End":"03:46.200","Text":"That\u0027s the obligation I owed you and that concludes Part a of the exercise."}],"ID":28692},{"Watched":false,"Name":"Exercise 9 - Part b","Duration":"40s","ChapterTopicVideoID":27506,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.000","Text":"Now we come to Part B of the exercise."},{"Start":"00:03.000 ","End":"00:08.985","Text":"Does the Sequence f_n converge uniformly on the interval minus Pi to Pi?"},{"Start":"00:08.985 ","End":"00:12.315","Text":"Now, remember the answer to Part A."},{"Start":"00:12.315 ","End":"00:17.385","Text":"Part A asks, does the sequence converge in the norm of L^2 for this interval?"},{"Start":"00:17.385 ","End":"00:19.770","Text":"We got the answer, no,"},{"Start":"00:19.770 ","End":"00:25.560","Text":"because uniform convergence implies convergence in the norm of L^2,"},{"Start":"00:25.560 ","End":"00:30.510","Text":"at least it does on a compact interval meaning closed and finite."},{"Start":"00:30.510 ","End":"00:33.240","Text":"This convergence implies this convergence."},{"Start":"00:33.240 ","End":"00:37.950","Text":"Then the non-convergence of this implies the non-convergence of this."},{"Start":"00:37.950 ","End":"00:41.050","Text":"That concludes Part B."}],"ID":28693},{"Watched":false,"Name":"Exercise 9 - Part c","Duration":"2m 20s","ChapterTopicVideoID":27507,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.445","Text":"Now we come to part c,"},{"Start":"00:02.445 ","End":"00:04.425","Text":"which is a bit like part b,"},{"Start":"00:04.425 ","End":"00:06.420","Text":"except that instead of f_n,"},{"Start":"00:06.420 ","End":"00:07.830","Text":"we use h_n,"},{"Start":"00:07.830 ","End":"00:10.215","Text":"which is the integral of f_n."},{"Start":"00:10.215 ","End":"00:17.669","Text":"Let\u0027s see. h_n is this integral and we\u0027ll actually compute the integral."},{"Start":"00:17.669 ","End":"00:24.555","Text":"We can bring out the sum outside the integral because it\u0027s a finite sum."},{"Start":"00:24.555 ","End":"00:29.435","Text":"We can also take the 1 over root k in front of the integral."},{"Start":"00:29.435 ","End":"00:33.800","Text":"It\u0027s easy to check that the integral of sine(kt) is 1 minus"},{"Start":"00:33.800 ","End":"00:41.255","Text":"cosine(kx) over k. Now we can bring the k out and we have k root k,"},{"Start":"00:41.255 ","End":"00:44.450","Text":"which is k^3 over 2,"},{"Start":"00:44.450 ","End":"00:49.370","Text":"or 1.5 and that\u0027s h_n(x) in terms of"},{"Start":"00:49.370 ","End":"00:54.530","Text":"a series of finite series 1 to n. Now if h_n converges,"},{"Start":"00:54.530 ","End":"00:58.160","Text":"then it converges to the infinite sum."},{"Start":"00:58.160 ","End":"01:00.020","Text":"When you let n go to infinity,"},{"Start":"01:00.020 ","End":"01:04.475","Text":"we just have to replace this n by infinity."},{"Start":"01:04.475 ","End":"01:07.805","Text":"The question is, does this series converge"},{"Start":"01:07.805 ","End":"01:11.000","Text":"uniformly by the definition of uniform convergence,"},{"Start":"01:11.000 ","End":"01:14.690","Text":"it\u0027s the same as asking whether the supremum of h"},{"Start":"01:14.690 ","End":"01:18.455","Text":"minus h_n goes to 0 as n goes to infinity,"},{"Start":"01:18.455 ","End":"01:23.795","Text":"but it\u0027s the supremum of the infinite sum minus the finite sum."},{"Start":"01:23.795 ","End":"01:26.705","Text":"That gives us the tail of the series."},{"Start":"01:26.705 ","End":"01:30.665","Text":"Here it\u0027s from n plus 1 to infinity."},{"Start":"01:30.665 ","End":"01:32.570","Text":"Subtract 1 to n,"},{"Start":"01:32.570 ","End":"01:35.915","Text":"and we have just from n plus 1 to infinity."},{"Start":"01:35.915 ","End":"01:40.010","Text":"Then this is less than or equal to 1 minus"},{"Start":"01:40.010 ","End":"01:44.769","Text":"cosine(x) is less than or equal to 2 in absolute value."},{"Start":"01:44.769 ","End":"01:48.710","Text":"This series goes to 0."},{"Start":"01:48.710 ","End":"01:51.800","Text":"The tail of the series, I\u0027ll explain."},{"Start":"01:51.800 ","End":"01:55.460","Text":"This is the tail of a convergent series,"},{"Start":"01:55.460 ","End":"01:59.540","Text":"2 over 1 over k^1 over 2."},{"Start":"01:59.540 ","End":"02:02.660","Text":"I might ask why is this series convergent?"},{"Start":"02:02.660 ","End":"02:04.460","Text":"It\u0027s a p-series."},{"Start":"02:04.460 ","End":"02:11.705","Text":"Sum 1 over k^p converges if p is bigger than 1."},{"Start":"02:11.705 ","End":"02:14.285","Text":"If this has a finite sum,"},{"Start":"02:14.285 ","End":"02:21.600","Text":"then the tail goes to 0 until the n to part c is yes, it converges uniformly."}],"ID":28694},{"Watched":false,"Name":"Exercise 9 - Part d","Duration":"37s","ChapterTopicVideoID":27508,"CourseChapterTopicPlaylistID":274167,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.215","Text":"Now part D, does the sequence h_n"},{"Start":"00:04.215 ","End":"00:09.750","Text":"converge in the norm of L^2 over the interval minus Pi, Pi."},{"Start":"00:09.750 ","End":"00:13.500","Text":"Now, remember that the answer to part C was yes."},{"Start":"00:13.500 ","End":"00:15.870","Text":"It converges uniformly,"},{"Start":"00:15.870 ","End":"00:19.785","Text":"and this is a finite closed interval, i e compact,"},{"Start":"00:19.785 ","End":"00:24.329","Text":"and when a sequence converges uniformly on a compact interval,"},{"Start":"00:24.329 ","End":"00:27.660","Text":"then it also converges in the norm of L^1,"},{"Start":"00:27.660 ","End":"00:29.400","Text":"L^2, most things."},{"Start":"00:29.400 ","End":"00:31.485","Text":"The answer is yes,"},{"Start":"00:31.485 ","End":"00:34.710","Text":"because uniform convergence implies convergence in L^2,"},{"Start":"00:34.710 ","End":"00:38.230","Text":"and we are done with this exercise."}],"ID":28695}],"Thumbnail":null,"ID":274167},{"Name":"Orthonormal Systems","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial - Orthogonal and Orthonormal Systems","Duration":"3m 1s","ChapterTopicVideoID":27477,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Starting a new topic,"},{"Start":"00:01.800 ","End":"00:05.100","Text":"orthogonal and orthonormal systems."},{"Start":"00:05.100 ","End":"00:07.260","Text":"Start with the orthogonal,"},{"Start":"00:07.260 ","End":"00:10.530","Text":"v is an inner product space and we\u0027ll only be talking about"},{"Start":"00:10.530 ","End":"00:15.075","Text":"infinite dimensional ones as all the function spaces are infinite dimensional,"},{"Start":"00:15.075 ","End":"00:20.175","Text":"and we have a infinite sequence of non-zero vectors."},{"Start":"00:20.175 ","End":"00:24.150","Text":"This sequence is called an infinite orthogonal system."},{"Start":"00:24.150 ","End":"00:27.060","Text":"Usually we drop the word infinite and it\u0027s understood"},{"Start":"00:27.060 ","End":"00:30.180","Text":"if the inner product of v_n with v_m"},{"Start":"00:30.180 ","End":"00:36.435","Text":"is 0 whenever n is not equal to m. An orthonormal system is similar."},{"Start":"00:36.435 ","End":"00:41.790","Text":"We start out with an orthogonal system that we add an extra condition,"},{"Start":"00:41.790 ","End":"00:45.690","Text":"basically that this is equal to1 if n is not equal"},{"Start":"00:45.690 ","End":"00:50.779","Text":"to m then it\u0027s called an orthonormal system."},{"Start":"00:50.779 ","End":"00:53.465","Text":"In an orthonormal system,"},{"Start":"00:53.465 ","End":"00:58.550","Text":"the inner product of v_n with v_m is 0 if n is not equal to m,"},{"Start":"00:58.550 ","End":"01:01.640","Text":"and 1 if n equals m. In orthogonal system,"},{"Start":"01:01.640 ","End":"01:03.425","Text":"we have the upper condition,"},{"Start":"01:03.425 ","End":"01:04.880","Text":"but not the lower."},{"Start":"01:04.880 ","End":"01:10.610","Text":"All we know is that if n equals m in the orthogonal case that this is non-zero,"},{"Start":"01:10.610 ","End":"01:13.860","Text":"because these are non-zero vectors."},{"Start":"01:13.990 ","End":"01:19.280","Text":"There are finite orthogonal systems and you could define it the same way,"},{"Start":"01:19.280 ","End":"01:22.940","Text":"just having a finite sequence of vectors,"},{"Start":"01:22.940 ","End":"01:25.175","Text":"but not in this course."},{"Start":"01:25.175 ","End":"01:27.710","Text":"Putting this in for completeness."},{"Start":"01:27.710 ","End":"01:31.520","Text":"We spoke about orthogonal where this inner product if n"},{"Start":"01:31.520 ","End":"01:35.300","Text":"equals m is not necessarily 1 but not 0."},{"Start":"01:35.300 ","End":"01:37.580","Text":"So what you can do if you want"},{"Start":"01:37.580 ","End":"01:40.455","Text":"an orthonormal system and you only have an orthogonal system,"},{"Start":"01:40.455 ","End":"01:43.055","Text":"you could divide each vector by a scalar,"},{"Start":"01:43.055 ","End":"01:44.690","Text":"which is the norm of this vector,"},{"Start":"01:44.690 ","End":"01:45.920","Text":"which is non-zero,"},{"Start":"01:45.920 ","End":"01:48.845","Text":"and then you get an orthonormal system."},{"Start":"01:48.845 ","End":"01:52.390","Text":"This is called normalizing the system."},{"Start":"01:52.390 ","End":"01:55.960","Text":"Now, a couple of definitions just for reference,"},{"Start":"01:55.960 ","End":"01:57.835","Text":"a closed orthonormal system."},{"Start":"01:57.835 ","End":"02:02.363","Text":"Let\u0027s say we have an orthogonal system, respectively orthonormal."},{"Start":"02:02.363 ","End":"02:05.980","Text":"We call it closed or closed orthogonal system,"},{"Start":"02:05.980 ","End":"02:07.959","Text":"if for every vector,"},{"Start":"02:07.959 ","End":"02:13.810","Text":"if the inner product of this vector with all the v_n is 0,"},{"Start":"02:13.810 ","End":"02:15.940","Text":"then the vector itself is 0."},{"Start":"02:15.940 ","End":"02:17.650","Text":"If this condition holds,"},{"Start":"02:17.650 ","End":"02:21.470","Text":"then it\u0027s called the closed orthogonal system, the v_n."},{"Start":"02:21.470 ","End":"02:27.060","Text":"Something similar, complete orthonormal system or orthogonal."},{"Start":"02:27.060 ","End":"02:30.460","Text":"This time we have same beginning as this,"},{"Start":"02:30.460 ","End":"02:31.764","Text":"just a different condition."},{"Start":"02:31.764 ","End":"02:33.415","Text":"Instead of this condition,"},{"Start":"02:33.415 ","End":"02:35.385","Text":"we have the following condition."},{"Start":"02:35.385 ","End":"02:40.283","Text":"That any v is the infinite linear combination of the v_n,"},{"Start":"02:40.283 ","End":"02:44.170","Text":"specifically the inner product of v with v_n times v_n."},{"Start":"02:44.170 ","End":"02:47.510","Text":"I want to say the infinite sum here,"},{"Start":"02:47.510 ","End":"02:52.280","Text":"I mean that it\u0027s the limit as n goes to infinity of the finite sum."},{"Start":"02:52.280 ","End":"03:01.980","Text":"This finite sum converges in the norm to v. That completes this introduction."}],"ID":28696},{"Watched":false,"Name":"Tutorial - Bessel\u0026#39;s Inequality","Duration":"3m 11s","ChapterTopicVideoID":27478,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.410","Text":"Next, we have an inequality related to orthonormal systems called Bessel\u0027s inequality."},{"Start":"00:07.410 ","End":"00:11.865","Text":"Let V be an inner product space and let the sequence e_k"},{"Start":"00:11.865 ","End":"00:16.830","Text":"be an orthonormal system in V. Then for all f in V,"},{"Start":"00:16.830 ","End":"00:20.880","Text":"we have this inequality that the infinite sum,"},{"Start":"00:20.880 ","End":"00:28.335","Text":"the series of the absolute value squared of inner product of f with e_k,"},{"Start":"00:28.335 ","End":"00:31.875","Text":"this is less than or equal to the norm of f squared."},{"Start":"00:31.875 ","End":"00:39.050","Text":"I\u0027ll just remark that if the orthonormal system happens to be complete,"},{"Start":"00:39.050 ","End":"00:45.545","Text":"then we get an equality here and it has a name Parseval\u0027s identity."},{"Start":"00:45.545 ","End":"00:48.145","Text":"Let\u0027s prove this."},{"Start":"00:48.145 ","End":"00:55.130","Text":"Let f belong to V and let n be natural number from 1 onwards."},{"Start":"00:55.130 ","End":"00:57.590","Text":"Then we have the following,"},{"Start":"00:57.590 ","End":"01:00.830","Text":"a norm of anything squared is bigger or equal to 0,"},{"Start":"01:00.830 ","End":"01:07.845","Text":"so I put here f minus the partial sum of the product of f with e_k times e_k,"},{"Start":"01:07.845 ","End":"01:11.200","Text":"sum from 1 to n squared."},{"Start":"01:11.200 ","End":"01:15.368","Text":"The norm squared is the inner product of the thing with itself,"},{"Start":"01:15.368 ","End":"01:18.710","Text":"but when we\u0027re going to multiply a sum with a sum,"},{"Start":"01:18.710 ","End":"01:19.910","Text":"we change the index."},{"Start":"01:19.910 ","End":"01:24.530","Text":"Here, I\u0027ve used j instead of k. Let\u0027s expand this."},{"Start":"01:24.530 ","End":"01:29.285","Text":"Each 1 of these with each 1 of these so we get 4 terms."},{"Start":"01:29.285 ","End":"01:31.975","Text":"First term is norm of f squared."},{"Start":"01:31.975 ","End":"01:35.344","Text":"In this we can bring the constant in front,"},{"Start":"01:35.344 ","End":"01:37.910","Text":"but we have to bring it as a conjugate."},{"Start":"01:37.910 ","End":"01:40.325","Text":"That\u0027s the way the inner product works."},{"Start":"01:40.325 ","End":"01:43.790","Text":"In this case, we don\u0027t need to make it conjugate."},{"Start":"01:43.790 ","End":"01:45.695","Text":"Here we get a double sum,"},{"Start":"01:45.695 ","End":"01:51.890","Text":"k and j and we get this brought in front, conjugated."},{"Start":"01:51.890 ","End":"01:56.240","Text":"Now because the e_k is an orthonormal system,"},{"Start":"01:56.240 ","End":"02:02.870","Text":"this inner product will be 0 except when j=k."},{"Start":"02:02.870 ","End":"02:06.500","Text":"The only thing we need to take from the blue sum is"},{"Start":"02:06.500 ","End":"02:10.550","Text":"the case where j=k and then we get e_k with e_k,"},{"Start":"02:10.550 ","End":"02:12.545","Text":"and that\u0027s 1 and we can drop it."},{"Start":"02:12.545 ","End":"02:17.480","Text":"What we\u0027re left with is this sum from k=1 to n,"},{"Start":"02:17.480 ","End":"02:20.615","Text":"f, e_k, f, e_k conjugate."},{"Start":"02:20.615 ","End":"02:25.819","Text":"Now this sum with a minus cancels with this sum with a plus."},{"Start":"02:25.819 ","End":"02:29.855","Text":"Bring this to the left-hand side and we\u0027ve got this."},{"Start":"02:29.855 ","End":"02:33.635","Text":"We can change the letter j back to k,"},{"Start":"02:33.635 ","End":"02:38.045","Text":"and then we get the absolute value squared of this."},{"Start":"02:38.045 ","End":"02:41.525","Text":"When you take a complex number and multiply it by its conjugate,"},{"Start":"02:41.525 ","End":"02:43.640","Text":"you get the absolute value squared."},{"Start":"02:43.640 ","End":"02:46.190","Text":"We\u0027re almost there, just that we have an n instead of"},{"Start":"02:46.190 ","End":"02:50.625","Text":"infinity so take the limit as n goes to infinity,"},{"Start":"02:50.625 ","End":"02:54.475","Text":"and then we just replace n with infinity."},{"Start":"02:54.475 ","End":"03:00.475","Text":"The partial sums are increasing and bounded above by this."},{"Start":"03:00.475 ","End":"03:03.055","Text":"When we let n go to infinity,"},{"Start":"03:03.055 ","End":"03:08.270","Text":"then it actually converges and the limit is also less than or equal to the upper bound."},{"Start":"03:08.270 ","End":"03:11.790","Text":"That\u0027s what we had to show and we\u0027re done."}],"ID":28697},{"Watched":false,"Name":"Exercise 1","Duration":"4m 18s","ChapterTopicVideoID":27479,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.790","Text":"In this exercise, V is the space L_2 on the closed interval minus Pi,"},{"Start":"00:05.790 ","End":"00:13.630","Text":"Pi of complex valued functions with the inner product as follows."},{"Start":"00:13.880 ","End":"00:21.420","Text":"This space L_2 on this interval means the square integrally functions or if you like,"},{"Start":"00:21.420 ","End":"00:24.390","Text":"the integral of the absolute value squared of the function"},{"Start":"00:24.390 ","End":"00:28.035","Text":"on this interval is less than infinity, it\u0027s finite."},{"Start":"00:28.035 ","End":"00:31.620","Text":"Now, let Phi_n,"},{"Start":"00:31.620 ","End":"00:33.620","Text":"not the individual Phi_n,"},{"Start":"00:33.620 ","End":"00:40.480","Text":"and the sequence Phi_n be defined by Phi_n of x equals cosine nx,"},{"Start":"00:40.480 ","End":"00:41.980","Text":"I should say n equals 1,"},{"Start":"00:41.980 ","End":"00:43.005","Text":"2, 3, 4, 5,"},{"Start":"00:43.005 ","End":"00:46.595","Text":"etc, positive natural numbers."},{"Start":"00:46.595 ","End":"00:52.630","Text":"We have to prove that the set Phi_n is an orthonormal system."},{"Start":"00:53.120 ","End":"00:57.050","Text":"Note by the way, that Phi_n really is in L_2,"},{"Start":"00:57.050 ","End":"01:00.170","Text":"the continuous on the compact interval."},{"Start":"01:00.170 ","End":"01:04.675","Text":"They are bounded and so square integrable, that\u0027s fine."},{"Start":"01:04.675 ","End":"01:08.720","Text":"What we need to show is the orthonormality,"},{"Start":"01:08.720 ","End":"01:12.500","Text":"which means that they\u0027re orthogonal to each other."},{"Start":"01:12.500 ","End":"01:16.825","Text":"If n is not equal to m, inner product is 0."},{"Start":"01:16.825 ","End":"01:21.850","Text":"They are normalized meaning Phi_n with Phi_m is 1."},{"Start":"01:21.940 ","End":"01:28.145","Text":"Let\u0027s take the case where n is not equal to m and see if we can get 0 out of this."},{"Start":"01:28.145 ","End":"01:33.020","Text":"Phi_n product with Phi_m by the definition of"},{"Start":"01:33.020 ","End":"01:40.170","Text":"the inner product here it is, is the following."},{"Start":"01:40.170 ","End":"01:43.520","Text":"Cosine nx times cosine mx."},{"Start":"01:43.520 ","End":"01:46.570","Text":"We don\u0027t really need the conjugate because this is real."},{"Start":"01:46.570 ","End":"01:48.990","Text":"This is equal to."},{"Start":"01:48.990 ","End":"01:52.985","Text":"Now we can use the trigonometric identity,"},{"Start":"01:52.985 ","End":"02:02.565","Text":"the one written here and apply it with Alpha is nx and Beta is mx."},{"Start":"02:02.565 ","End":"02:07.175","Text":"Then we get using this formula the following."},{"Start":"02:07.175 ","End":"02:10.715","Text":"The 2 here is the 2 here."},{"Start":"02:10.715 ","End":"02:17.300","Text":"This is equal to the integral of cosine n minus mx is sine n minus x"},{"Start":"02:17.300 ","End":"02:24.700","Text":"divided by n minus m. The reason I\u0027ve highlighted this is to note that this is not 0,"},{"Start":"02:24.700 ","End":"02:27.030","Text":"n is not equal to m,"},{"Start":"02:27.030 ","End":"02:29.130","Text":"so this is not 0,"},{"Start":"02:29.130 ","End":"02:37.210","Text":"and this is not 0 either because n and m are positive integers."},{"Start":"02:37.210 ","End":"02:39.890","Text":"If we were talking about all integers,"},{"Start":"02:39.890 ","End":"02:44.450","Text":"we\u0027d have to make sure that n is not equal to plus or minus m. Anyway."},{"Start":"02:44.450 ","End":"02:53.790","Text":"What we get is if you substitute x equals Pi or minus Pi in either of these we get 0,"},{"Start":"02:54.020 ","End":"02:57.410","Text":"because the sine of any whole number"},{"Start":"02:57.410 ","End":"03:06.670","Text":"times x of any whole number times Pi or minus Pi is 0."},{"Start":"03:07.010 ","End":"03:11.485","Text":"We get 0 here and here."},{"Start":"03:11.485 ","End":"03:15.275","Text":"It\u0027s actually 0 minus 0 because 0,"},{"Start":"03:15.275 ","End":"03:18.260","Text":"when we substitute Pi is 0 and we substitute minus Pi is 0,"},{"Start":"03:18.260 ","End":"03:20.540","Text":"minus 0 is 0, similarly here."},{"Start":"03:20.540 ","End":"03:24.080","Text":"Anyway, we\u0027ll boils down to 0. That\u0027s what we wanted to show."},{"Start":"03:24.080 ","End":"03:27.920","Text":"That\u0027s just the first part where n is not equal to m. Now we want"},{"Start":"03:27.920 ","End":"03:32.600","Text":"the other case for n is equal to m. Then we\u0027ll use a different trigonometric formula."},{"Start":"03:32.600 ","End":"03:35.630","Text":"What we get is this integral cosine nx cosine nx,"},{"Start":"03:35.630 ","End":"03:37.160","Text":"which makes it squared."},{"Start":"03:37.160 ","End":"03:39.290","Text":"Then we\u0027ll use this formula."},{"Start":"03:39.290 ","End":"03:41.225","Text":"If we use this formula,"},{"Start":"03:41.225 ","End":"03:44.420","Text":"then we get this."},{"Start":"03:44.420 ","End":"03:52.489","Text":"The integral of this is x plus sine 2nx over 2n, 2n is not 0."},{"Start":"03:52.489 ","End":"03:55.550","Text":"Again, the sine of"},{"Start":"03:55.550 ","End":"04:03.225","Text":"2nx will be 0 when x equals Pi or minus Pi,"},{"Start":"04:03.225 ","End":"04:04.830","Text":"0 minus 0 is 0."},{"Start":"04:04.830 ","End":"04:08.445","Text":"We just have the x bit from Pi to minus Pi,"},{"Start":"04:08.445 ","End":"04:09.765","Text":"minus Pi to Pi."},{"Start":"04:09.765 ","End":"04:12.285","Text":"Anyway, we get Pi minus minus Pi,"},{"Start":"04:12.285 ","End":"04:15.090","Text":"which is 2Pi over Pi, which is 1."},{"Start":"04:15.090 ","End":"04:16.425","Text":"That\u0027s what we needed,"},{"Start":"04:16.425 ","End":"04:19.300","Text":"and that concludes this exercise."}],"ID":28698},{"Watched":false,"Name":"Exercise 2","Duration":"2m 45s","ChapterTopicVideoID":27480,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.640","Text":"In this exercise, V is the space, well,"},{"Start":"00:03.640 ","End":"00:06.880","Text":"this is square integral functions on the interval 0"},{"Start":"00:06.880 ","End":"00:11.755","Text":"Pi and the inner product is defined as follows."},{"Start":"00:11.755 ","End":"00:18.520","Text":"We have a set Phi n functions"},{"Start":"00:18.520 ","End":"00:22.690","Text":"defined for each n by Phi n^x is sine nx."},{"Start":"00:22.690 ","End":"00:25.510","Text":"We have to prove that the set Phi n,"},{"Start":"00:25.510 ","End":"00:27.730","Text":"where n goes from 1 to infinity is"},{"Start":"00:27.730 ","End":"00:33.310","Text":"an orthonormal system test to remark that the Phi n really are in"},{"Start":"00:33.310 ","End":"00:37.060","Text":"this space because they\u0027re continuous and a compact interval therefore"},{"Start":"00:37.060 ","End":"00:42.815","Text":"bounded on the compact interval and so square integral."},{"Start":"00:42.815 ","End":"00:49.520","Text":"What we need to show for the orthonormal system is that the product of two of them,"},{"Start":"00:49.520 ","End":"00:54.800","Text":"Phi n with Phi m is 0 if n is not equal to m and 1 if n=m,"},{"Start":"00:54.800 ","End":"00:59.360","Text":"I will start with the case where n is not equal to m and we want to get 0."},{"Start":"00:59.360 ","End":"01:03.469","Text":"The inner product using the definition is this."},{"Start":"01:03.469 ","End":"01:06.620","Text":"We don\u0027t need the conjugate because real numbers,"},{"Start":"01:06.620 ","End":"01:09.230","Text":"the sign of the real is real."},{"Start":"01:09.230 ","End":"01:14.060","Text":"This is equal to use a trigonometric identity,"},{"Start":"01:14.060 ","End":"01:19.115","Text":"this one, Alpha is nx and Beta is mx and substitute."},{"Start":"01:19.115 ","End":"01:20.855","Text":"We get this."},{"Start":"01:20.855 ","End":"01:25.095","Text":"Then the integral of cosine is sine,"},{"Start":"01:25.095 ","End":"01:28.880","Text":"but we have to divide by the inner derivative n minus m and here n"},{"Start":"01:28.880 ","End":"01:33.140","Text":"plus m. Now n is not equal to m and n and m are positive number,"},{"Start":"01:33.140 ","End":"01:36.260","Text":"so this is not 0 and this is not 0 either."},{"Start":"01:36.260 ","End":"01:42.695","Text":"But when we plug in x=Pi or 0 to sine(kx),"},{"Start":"01:42.695 ","End":"01:45.310","Text":"then we get 0."},{"Start":"01:45.310 ","End":"01:48.710","Text":"So this just becomes 0."},{"Start":"01:48.710 ","End":"01:53.599","Text":"Each of these, and therefore the whole thing is 0 and that\u0027s what we wanted."},{"Start":"01:53.599 ","End":"01:57.545","Text":"That\u0027s for the orthogonal when they are not equal."},{"Start":"01:57.545 ","End":"02:00.695","Text":"Now the other case where n=m,"},{"Start":"02:00.695 ","End":"02:03.260","Text":"then what we get instead of sine,"},{"Start":"02:03.260 ","End":"02:05.770","Text":"sine, we get sine squared."},{"Start":"02:05.770 ","End":"02:08.450","Text":"Again, we use the trigonometric identity,"},{"Start":"02:08.450 ","End":"02:10.070","Text":"we use this one."},{"Start":"02:10.070 ","End":"02:11.630","Text":"If we do that,"},{"Start":"02:11.630 ","End":"02:15.920","Text":"then we get the integral of 1 minus cosine(2nx)."},{"Start":"02:15.920 ","End":"02:20.630","Text":"This integral is x plus sine 2nx over 2n."},{"Start":"02:20.630 ","End":"02:26.890","Text":"Now, again, the sign of a whole number of Pi is always 0,"},{"Start":"02:26.890 ","End":"02:30.760","Text":"so we get 0 minus 0 is 0 here."},{"Start":"02:30.760 ","End":"02:33.415","Text":"We just have the bit from the x,"},{"Start":"02:33.415 ","End":"02:37.170","Text":"and the x gives us Pi minus 0."},{"Start":"02:37.170 ","End":"02:40.365","Text":"We get Pi over Pi, which is 1,"},{"Start":"02:40.365 ","End":"02:45.340","Text":"and that\u0027s the second case and that completes this exercise."}],"ID":28699},{"Watched":false,"Name":"Exercise 3","Duration":"2m 11s","ChapterTopicVideoID":27481,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.540","Text":"This is an important exercise because we will refer to"},{"Start":"00:03.540 ","End":"00:07.350","Text":"its result in some of the future exercises,"},{"Start":"00:07.350 ","End":"00:09.330","Text":"that V be an inner product space,"},{"Start":"00:09.330 ","End":"00:16.980","Text":"and let phi n be a subset of v and n goes from zero to infinity such that each phi n(x),"},{"Start":"00:16.980 ","End":"00:20.805","Text":"it\u0027s a polynomial of degree n. It has the property"},{"Start":"00:20.805 ","End":"00:25.590","Text":"that the inner product of phi n(x) with x to the m is zero,"},{"Start":"00:25.590 ","End":"00:33.345","Text":"whenever m is less than n. We have to prove that phi n forms an orthogonal set."},{"Start":"00:33.345 ","End":"00:36.540","Text":"Really an orthogonal system is what I mean to say in"},{"Start":"00:36.540 ","End":"00:41.220","Text":"V. Now we just require it to be orthogonal, not necessarily orthonormal."},{"Start":"00:41.220 ","End":"00:45.620","Text":"All we have to show is that whenever n is different from m,"},{"Start":"00:45.620 ","End":"00:49.205","Text":"that the inner product phi n with phi m is 0."},{"Start":"00:49.205 ","End":"00:51.845","Text":"Now, we can also assume,"},{"Start":"00:51.845 ","End":"00:55.070","Text":"we say in mathematics without loss of generality that m"},{"Start":"00:55.070 ","End":"00:58.715","Text":"is less than n. Because if m is bigger than n,"},{"Start":"00:58.715 ","End":"01:00.290","Text":"then you can just flip them around,"},{"Start":"01:00.290 ","End":"01:01.610","Text":"rename them call n,"},{"Start":"01:01.610 ","End":"01:03.230","Text":"m and call m, n anyway,"},{"Start":"01:03.230 ","End":"01:04.940","Text":"you can always make sure that it\u0027s m,"},{"Start":"01:04.940 ","End":"01:06.740","Text":"that is the smaller one of the two,"},{"Start":"01:06.740 ","End":"01:08.805","Text":"and you don\u0027t lose any generality."},{"Start":"01:08.805 ","End":"01:12.755","Text":"Phi m(x) is a polynomial of degree m,"},{"Start":"01:12.755 ","End":"01:15.485","Text":"so it\u0027s of this form."},{"Start":"01:15.485 ","End":"01:20.540","Text":"Phi n with phi m is phi n with,"},{"Start":"01:20.540 ","End":"01:24.260","Text":"instead of phi m put this polynomial here and then use"},{"Start":"01:24.260 ","End":"01:28.280","Text":"linearity to break it up as follows."},{"Start":"01:28.280 ","End":"01:30.830","Text":"We can also take the scalar,"},{"Start":"01:30.830 ","End":"01:32.270","Text":"the constant in front,"},{"Start":"01:32.270 ","End":"01:36.770","Text":"but it has to come out with a conjugate sign, like so."},{"Start":"01:36.770 ","End":"01:41.195","Text":"Now I claim that each of these inner product is 0,"},{"Start":"01:41.195 ","End":"01:47.075","Text":"because we know that phi n with any xk,"},{"Start":"01:47.075 ","End":"01:48.185","Text":"use a different letter,"},{"Start":"01:48.185 ","End":"01:55.235","Text":"is zero whenever k is less than n. It is less than n because k is one of these,"},{"Start":"01:55.235 ","End":"01:59.450","Text":"and each of these is less than or equal to m and m is less than n. Any one of these m,"},{"Start":"01:59.450 ","End":"02:01.325","Text":"m minus 1 down to 1,0,"},{"Start":"02:01.325 ","End":"02:03.345","Text":"they\u0027re all less than n,"},{"Start":"02:03.345 ","End":"02:06.255","Text":"so phi n with all of these is 0,"},{"Start":"02:06.255 ","End":"02:07.845","Text":"which shows that phi n,"},{"Start":"02:07.845 ","End":"02:09.495","Text":"phi m is 0,"},{"Start":"02:09.495 ","End":"02:11.980","Text":"and we are done."}],"ID":28700},{"Watched":false,"Name":"Exercise 4","Duration":"2m 12s","ChapterTopicVideoID":27482,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.420","Text":"In this exercise, we take v to be the space of"},{"Start":"00:03.420 ","End":"00:09.945","Text":"piecewise continuous functions on the interval from e to the minus Pi to e to the Pi."},{"Start":"00:09.945 ","End":"00:13.485","Text":"The inner product is the following."},{"Start":"00:13.485 ","End":"00:16.260","Text":"It\u0027s the integral of this with this conjugate."},{"Start":"00:16.260 ","End":"00:18.510","Text":"This is called the weight function."},{"Start":"00:18.510 ","End":"00:21.810","Text":"Notice that 1/x is not 0 on this interval."},{"Start":"00:21.810 ","End":"00:24.029","Text":"This is a strictly positive interval."},{"Start":"00:24.029 ","End":"00:25.680","Text":"We\u0027re okay there."},{"Start":"00:25.680 ","End":"00:33.795","Text":"Now we define Phi n of x a set for all n to be sin(n ln x)."},{"Start":"00:33.795 ","End":"00:36.570","Text":"We have to prove that the set Phi n,"},{"Start":"00:36.570 ","End":"00:40.160","Text":"where n goes from 1 to infinity is an orthogonal system in"},{"Start":"00:40.160 ","End":"00:45.950","Text":"v. Basically all we need to show is that Phi n inner product with Phi m is 0"},{"Start":"00:45.950 ","End":"00:51.080","Text":"whenever n is not equal to m. This inner product is equal to"},{"Start":"00:51.080 ","End":"00:56.795","Text":"the inner product of sin(n ln x), sine(m ln x)."},{"Start":"00:56.795 ","End":"00:59.435","Text":"Find the definition of the inner product."},{"Start":"00:59.435 ","End":"01:02.675","Text":"It\u0027s this with the 1/x here."},{"Start":"01:02.675 ","End":"01:05.015","Text":"Make a substitution."},{"Start":"01:05.015 ","End":"01:08.660","Text":"We let t be ln x."},{"Start":"01:08.660 ","End":"01:12.070","Text":"We have t here and we have t here."},{"Start":"01:12.070 ","End":"01:16.680","Text":"Also also dt is 1/x dx."},{"Start":"01:16.680 ","End":"01:19.230","Text":"Altogether this part comes out dt."},{"Start":"01:19.230 ","End":"01:22.495","Text":"This is sin(mt), sin(nt)."},{"Start":"01:22.495 ","End":"01:25.745","Text":"We also change the limits of integration."},{"Start":"01:25.745 ","End":"01:30.620","Text":"We need to take the limits for t. When x is from here to here,"},{"Start":"01:30.620 ","End":"01:32.240","Text":"t is the natural log of this,"},{"Start":"01:32.240 ","End":"01:34.540","Text":"which is from minus Pi to Pi."},{"Start":"01:34.540 ","End":"01:39.440","Text":"This is 0 when n is not equal to m because we shown this in the previous exercise"},{"Start":"01:39.440 ","End":"01:45.455","Text":"that sin(nt) is an orthogonal system on the interval from minus Pi to Pi."},{"Start":"01:45.455 ","End":"01:46.970","Text":"We\u0027re basically done."},{"Start":"01:46.970 ","End":"01:50.785","Text":"But if you want to see again the proof of this,"},{"Start":"01:50.785 ","End":"01:54.330","Text":"here it is from the other exercise."},{"Start":"01:54.330 ","End":"01:55.820","Text":"It says even more."},{"Start":"01:55.820 ","End":"02:01.450","Text":"It tells you what happens when m equals n or j equals k, you get Pi."},{"Start":"02:01.450 ","End":"02:06.305","Text":"The important thing here is that n is not equal to m,"},{"Start":"02:06.305 ","End":"02:08.135","Text":"and then this is not 0."},{"Start":"02:08.135 ","End":"02:10.280","Text":"Otherwise it doesn\u0027t work."},{"Start":"02:10.280 ","End":"02:13.140","Text":"Now we\u0027re really are done."}],"ID":28701},{"Watched":false,"Name":"Exercise 5","Duration":"5m 44s","ChapterTopicVideoID":27483,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:05.400","Text":"In this exercise we have the space L^2(a,b),"},{"Start":"00:05.400 ","End":"00:09.195","Text":"and we have an inner product like usual,"},{"Start":"00:09.195 ","End":"00:11.730","Text":"and we have a set Phi_n,"},{"Start":"00:11.730 ","End":"00:13.260","Text":"and equals 0 to infinity."},{"Start":"00:13.260 ","End":"00:16.380","Text":"This is a closed orthonormal set."},{"Start":"00:16.380 ","End":"00:18.135","Text":"I\u0027ll remind you in a moment."},{"Start":"00:18.135 ","End":"00:21.390","Text":"We label the inner product"},{"Start":"00:21.390 ","End":"00:24.000","Text":"with the 1 because in a moment we\u0027re going to have another one,"},{"Start":"00:24.000 ","End":"00:25.680","Text":"and we\u0027ll label that with a 2."},{"Start":"00:25.680 ","End":"00:29.010","Text":"Let c and d be positive real constants,"},{"Start":"00:29.010 ","End":"00:35.105","Text":"and define Psi_n(x) in terms of Phi_n(x) by this formula."},{"Start":"00:35.105 ","End":"00:43.100","Text":"We have to prove that the set Psi_n is a closed orthonormal set from a different space,"},{"Start":"00:43.100 ","End":"00:45.185","Text":"as base L^2 on a different interval,"},{"Start":"00:45.185 ","End":"00:46.400","Text":"a minus d over c,"},{"Start":"00:46.400 ","End":"00:47.855","Text":"b minus d over c,"},{"Start":"00:47.855 ","End":"00:50.505","Text":"c is not 0, so we\u0027re okay there."},{"Start":"00:50.505 ","End":"00:55.400","Text":"The inner product on this space is the integral from a minus d over c,"},{"Start":"00:55.400 ","End":"00:56.690","Text":"to b minus d over c,"},{"Start":"00:56.690 ","End":"01:00.335","Text":"also of f(x), g(x) conjugate dx."},{"Start":"01:00.335 ","End":"01:05.360","Text":"Now I\u0027ll remind you what it means for a system to be closed."},{"Start":"01:05.360 ","End":"01:08.090","Text":"This orthonormal set or orthogonal set,"},{"Start":"01:08.090 ","End":"01:09.950","Text":"is closed in V,"},{"Start":"01:09.950 ","End":"01:14.240","Text":"means that if the inner product of v with all the v_n is 0,"},{"Start":"01:14.240 ","End":"01:17.799","Text":"then necessarily v is the 0 vector."},{"Start":"01:17.799 ","End":"01:19.670","Text":"We have 2 things to show."},{"Start":"01:19.670 ","End":"01:23.750","Text":"We have to show that Psi_n is orthonormal,"},{"Start":"01:23.750 ","End":"01:27.260","Text":"and we also have to show that it\u0027s closed."},{"Start":"01:27.260 ","End":"01:31.730","Text":"Let\u0027s compute the inner product of Psi_n with Psi_m,"},{"Start":"01:31.730 ","End":"01:34.580","Text":"and show that that is 0,"},{"Start":"01:34.580 ","End":"01:36.065","Text":"if n is not equal to m,"},{"Start":"01:36.065 ","End":"01:41.000","Text":"and 1 if n is equal to m. This is equal to, by the definition,"},{"Start":"01:41.000 ","End":"01:44.600","Text":"this, and he\u0027s reminder that Psi_n,"},{"Start":"01:44.600 ","End":"01:47.335","Text":"this is the formula for it in terms of Phi_n."},{"Start":"01:47.335 ","End":"01:53.435","Text":"What we get then just replacing Psi with its definition, we get this."},{"Start":"01:53.435 ","End":"02:00.025","Text":"Now we\u0027re going to make a obvious substitution to let t equals cx plus d,"},{"Start":"02:00.025 ","End":"02:03.960","Text":"and then dt will equal c times dx."},{"Start":"02:03.960 ","End":"02:09.244","Text":"After the substitution we\u0027ll get from this,"},{"Start":"02:09.244 ","End":"02:12.335","Text":"note that we have square root of c,"},{"Start":"02:12.335 ","End":"02:16.550","Text":"and square root of c and dx, colored these,"},{"Start":"02:16.550 ","End":"02:22.130","Text":"and if you multiply them together we get c dx, which is dt."},{"Start":"02:22.130 ","End":"02:24.545","Text":"That\u0027s where this dt comes from."},{"Start":"02:24.545 ","End":"02:26.300","Text":"This is just Phi_n(t),"},{"Start":"02:26.300 ","End":"02:29.970","Text":"and this is Phi_m(t) conjugate."},{"Start":"02:29.970 ","End":"02:33.725","Text":"Also note that when we let t equals cx plus d,"},{"Start":"02:33.725 ","End":"02:36.210","Text":"t goes from a to b."},{"Start":"02:36.210 ","End":"02:40.115","Text":"Now we know that Phi_n is an orthonormal system,"},{"Start":"02:40.115 ","End":"02:44.225","Text":"and this is just the inner product of Phi_n with Phi_m."},{"Start":"02:44.225 ","End":"02:45.830","Text":"I skipped a stage."},{"Start":"02:45.830 ","End":"02:48.185","Text":"This is angular brackets Phi_n,"},{"Start":"02:48.185 ","End":"02:51.139","Text":"Phi_m, and therefore because it\u0027s orthonormal,"},{"Start":"02:51.139 ","End":"02:52.910","Text":"this is 0 if n is not equal to n,"},{"Start":"02:52.910 ","End":"02:57.485","Text":"and 1 if n equals m This reflects back to the Psi_n and Psi_m."},{"Start":"02:57.485 ","End":"03:00.740","Text":"This is also orthonormal on its interval."},{"Start":"03:00.740 ","End":"03:05.065","Text":"Next we have to show that the system is closed."},{"Start":"03:05.065 ","End":"03:09.275","Text":"Suppose we have a function in L^2 of this space,"},{"Start":"03:09.275 ","End":"03:14.375","Text":"and suppose it has the property that f inner product with Psi_n"},{"Start":"03:14.375 ","End":"03:19.620","Text":"is 0 for all n. We have to show that f is the 0 function."},{"Start":"03:19.620 ","End":"03:22.770","Text":"That\u0027s what the closed means."},{"Start":"03:22.770 ","End":"03:25.485","Text":"Suppose that for all n,"},{"Start":"03:25.485 ","End":"03:30.675","Text":"f with Psi_n is 0 in a product number 2."},{"Start":"03:30.675 ","End":"03:33.935","Text":"Then that means that for all n,"},{"Start":"03:33.935 ","End":"03:39.320","Text":"this integral, that\u0027s the definition of this inner product, this is 0."},{"Start":"03:39.320 ","End":"03:44.450","Text":"Then we can substitute what Psi_n is in terms of Phi_n."},{"Start":"03:44.450 ","End":"03:46.885","Text":"We get this expression."},{"Start":"03:46.885 ","End":"03:51.710","Text":"Then we can make a substitution like before,"},{"Start":"03:51.710 ","End":"03:55.970","Text":"t equals cx plus d, dt equals cdx."},{"Start":"03:55.970 ","End":"03:59.248","Text":"We just divide dx,"},{"Start":"03:59.248 ","End":"04:05.670","Text":"and we get this is dt over c. cx plus d is the t here,"},{"Start":"04:05.670 ","End":"04:09.620","Text":"and x, just bring the d over to the other side divided by cx,"},{"Start":"04:09.620 ","End":"04:13.775","Text":"is t minus d over c. This is the expression we get."},{"Start":"04:13.775 ","End":"04:16.205","Text":"Since we\u0027re comparing to 0,"},{"Start":"04:16.205 ","End":"04:17.630","Text":"we don\u0027t need these constants."},{"Start":"04:17.630 ","End":"04:20.830","Text":"You can throw this one out and you can throw that one out."},{"Start":"04:20.830 ","End":"04:23.340","Text":"That simplifies it a little bit."},{"Start":"04:23.340 ","End":"04:28.720","Text":"Then this is just the inner product of f(t) minus d over c,"},{"Start":"04:28.720 ","End":"04:30.835","Text":"with Phi_n of t,"},{"Start":"04:30.835 ","End":"04:32.395","Text":"the inner product number 1."},{"Start":"04:32.395 ","End":"04:35.040","Text":"For all of that this is 0."},{"Start":"04:35.040 ","End":"04:38.960","Text":"This is a function of t. Maybe I should have relabeled"},{"Start":"04:38.960 ","End":"04:42.245","Text":"it that f(t) minus d over c is g(t)."},{"Start":"04:42.245 ","End":"04:47.930","Text":"Anyway, it\u0027s a function of t defined through f. That means"},{"Start":"04:47.930 ","End":"04:54.470","Text":"that because Phi_n is closed that this function as a function of t is 0,"},{"Start":"04:54.470 ","End":"04:58.065","Text":"and t is in the interval a, b,"},{"Start":"04:58.065 ","End":"05:01.235","Text":"just like we converted the limits of the integral here,"},{"Start":"05:01.235 ","End":"05:03.530","Text":"t is cx plus d,"},{"Start":"05:03.530 ","End":"05:05.390","Text":"and if x goes from here to here,"},{"Start":"05:05.390 ","End":"05:07.130","Text":"then t goes from here to here."},{"Start":"05:07.130 ","End":"05:09.295","Text":"This f is a function of t,"},{"Start":"05:09.295 ","End":"05:11.650","Text":"t is defined in a, b,"},{"Start":"05:11.650 ","End":"05:15.560","Text":"and t minus d over c is defined on this interval,"},{"Start":"05:15.560 ","End":"05:18.915","Text":"from a minus d over c to b minus d over c. Anyway."},{"Start":"05:18.915 ","End":"05:23.000","Text":"Now let x belongs to this interval."},{"Start":"05:23.000 ","End":"05:27.080","Text":"Then cx plus d belongs to the interval a, b,"},{"Start":"05:27.080 ","End":"05:30.775","Text":"call it t. f(x),"},{"Start":"05:30.775 ","End":"05:33.675","Text":"which is f(t) minus d over c, is 0."},{"Start":"05:33.675 ","End":"05:35.580","Text":"For all x in this interval,"},{"Start":"05:35.580 ","End":"05:40.780","Text":"f(x) is 0, which means that f is the 0 function of this interval."},{"Start":"05:40.780 ","End":"05:43.140","Text":"That\u0027s what we wanted to show,"},{"Start":"05:43.140 ","End":"05:45.670","Text":"and now we\u0027re done."}],"ID":28702},{"Watched":false,"Name":"Exercise 6","Duration":"1m 41s","ChapterTopicVideoID":27469,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.920","Text":"In this exercise, V is an inner product space complex,"},{"Start":"00:04.920 ","End":"00:11.175","Text":"and we have a finite set of orthonormal vectors in this space."},{"Start":"00:11.175 ","End":"00:14.760","Text":"We have to prove that for any vector in V,"},{"Start":"00:14.760 ","End":"00:17.325","Text":"we have the following equality."},{"Start":"00:17.325 ","End":"00:23.115","Text":"Remember orthonormal means that the following condition holds for the e_i."},{"Start":"00:23.115 ","End":"00:25.980","Text":"Now, let\u0027s start computing the left-hand side."},{"Start":"00:25.980 ","End":"00:31.345","Text":"The norm squared is the inner product of what it is with itself."},{"Start":"00:31.345 ","End":"00:34.710","Text":"Because we have sum with n here,"},{"Start":"00:34.710 ","End":"00:36.450","Text":"we need a sum with a different letter."},{"Start":"00:36.450 ","End":"00:40.650","Text":"Otherwise, we can\u0027t get product of one from here, one from here."},{"Start":"00:40.650 ","End":"00:45.050","Text":"Here we use n, here we use m. Now we can expand"},{"Start":"00:45.050 ","End":"00:51.095","Text":"the first component of the inner product as a sum and take the constants out."},{"Start":"00:51.095 ","End":"00:53.570","Text":"But when we do it with the second sum,"},{"Start":"00:53.570 ","End":"00:57.440","Text":"we need to take the constants out conjugated."},{"Start":"00:57.440 ","End":"00:58.790","Text":"We\u0027ve done this thing before,"},{"Start":"00:58.790 ","End":"01:00.560","Text":"so I won\u0027t go into much detail."},{"Start":"01:00.560 ","End":"01:03.345","Text":"Then, we know that e_n,"},{"Start":"01:03.345 ","End":"01:10.700","Text":"e_m is mostly zero except when m is equal to n. From all this sum,"},{"Start":"01:10.700 ","End":"01:14.510","Text":"we only take one of the sums, one of the terms,"},{"Start":"01:14.510 ","End":"01:19.940","Text":"the one where m is equal to n and that gives us e_n here and e_n here."},{"Start":"01:19.940 ","End":"01:22.985","Text":"But this disappears because e_n e_n is 1."},{"Start":"01:22.985 ","End":"01:24.320","Text":"Also, we don\u0027t have the sum,"},{"Start":"01:24.320 ","End":"01:28.890","Text":"we just had the n. We get this."},{"Start":"01:28.890 ","End":"01:32.900","Text":"Then v_n is a complex number times its conjugate,"},{"Start":"01:32.900 ","End":"01:35.705","Text":"that will give its absolute value squared."},{"Start":"01:35.705 ","End":"01:39.620","Text":"This is exactly what we had to prove that this is equal to this,"},{"Start":"01:39.620 ","End":"01:41.880","Text":"so we are done."}],"ID":28703},{"Watched":false,"Name":"Exercise 7 - Chebyshev Polynomials","Duration":"4m 57s","ChapterTopicVideoID":27470,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.570","Text":"In this exercise, we let K be the space of"},{"Start":"00:03.570 ","End":"00:06.180","Text":"piecewise continuous functions on"},{"Start":"00:06.180 ","End":"00:10.935","Text":"the open interval from minus 1 1 to the complex numbers,"},{"Start":"00:10.935 ","End":"00:14.115","Text":"and the inner product is defined as follows."},{"Start":"00:14.115 ","End":"00:15.900","Text":"It starts out as usual,"},{"Start":"00:15.900 ","End":"00:18.690","Text":"with f times g conjugate,"},{"Start":"00:18.690 ","End":"00:21.465","Text":"then times something called the weight function."},{"Start":"00:21.465 ","End":"00:22.620","Text":"You put anything here,"},{"Start":"00:22.620 ","End":"00:24.315","Text":"it\u0027s a weight function."},{"Start":"00:24.315 ","End":"00:28.440","Text":"This is well-defined because from minus 1 to 1,"},{"Start":"00:28.440 ","End":"00:33.650","Text":"x^2 is less than 1 so the denominator is defined,"},{"Start":"00:33.650 ","End":"00:36.155","Text":"it\u0027s square root of a positive number."},{"Start":"00:36.155 ","End":"00:39.680","Text":"Now we define the Chebyshev polynomials."},{"Start":"00:39.680 ","End":"00:45.915","Text":"T_n(x) is equal to cos(x times arc cosine(x))."},{"Start":"00:45.915 ","End":"00:50.360","Text":"Arc cosine is sometimes written as cosine to the power of minus 1."},{"Start":"00:50.360 ","End":"00:56.800","Text":"The letter T comes from an alternative spelling of Chebyshev that begins with TCH,"},{"Start":"00:56.800 ","End":"01:00.380","Text":"might see Tchaikovsky spelled with the TCH and so on."},{"Start":"01:00.380 ","End":"01:05.450","Text":"We have to prove that the set of T_n and,"},{"Start":"01:05.450 ","End":"01:06.950","Text":"n goes from 0 to infinity,"},{"Start":"01:06.950 ","End":"01:11.810","Text":"is an orthogonal system and we have to find the normalizing constants,"},{"Start":"01:11.810 ","End":"01:14.725","Text":"in other words the constants Alpha n,"},{"Start":"01:14.725 ","End":"01:18.410","Text":"such that if you multiply Alpha n times T_n,"},{"Start":"01:18.410 ","End":"01:21.185","Text":"then you get an orthonormal system."},{"Start":"01:21.185 ","End":"01:25.280","Text":"It\u0027s not obvious that the T_n are polynomials in this definition,"},{"Start":"01:25.280 ","End":"01:26.870","Text":"but in fact they are."},{"Start":"01:26.870 ","End":"01:30.155","Text":"Here\u0027s a list of the first 6 of them."},{"Start":"01:30.155 ","End":"01:38.060","Text":"Each polynomial T_n has a degree n. Where the cosine comes in,"},{"Start":"01:38.060 ","End":"01:39.908","Text":"won\u0027t go into detail,"},{"Start":"01:39.908 ","End":"01:43.640","Text":"but if you look at this table of cosine of a multiple angle,"},{"Start":"01:43.640 ","End":"01:46.560","Text":"for example, look at cos(3x),"},{"Start":"01:46.560 ","End":"01:52.260","Text":"full cos(x^3) minus 3 cos(x) and that\u0027s similar to T_3,"},{"Start":"01:52.260 ","End":"01:54.525","Text":"which is 4x^3 minus 3x."},{"Start":"01:54.525 ","End":"01:57.530","Text":"If you think about it, this definition exactly"},{"Start":"01:57.530 ","End":"02:01.700","Text":"says that you take cosine of n times an angle,"},{"Start":"02:01.700 ","End":"02:08.465","Text":"and then replace cosine x with x and you get the corresponding Chebyshev polynomial."},{"Start":"02:08.465 ","End":"02:10.085","Text":"That was just for interest\u0027s sake."},{"Start":"02:10.085 ","End":"02:13.070","Text":"Yeah, like I just said, T_n is a polynomial of"},{"Start":"02:13.070 ","End":"02:16.640","Text":"degree n. What we need to show first of all,"},{"Start":"02:16.640 ","End":"02:25.335","Text":"is the orthogonality is that T_n in a product with T_m is 0 if n is not equal to m. T_n,"},{"Start":"02:25.335 ","End":"02:29.604","Text":"T_m is this integral with the weight function."},{"Start":"02:29.604 ","End":"02:31.638","Text":"This is equal to,"},{"Start":"02:31.638 ","End":"02:33.240","Text":"from the definition here,"},{"Start":"02:33.240 ","End":"02:37.190","Text":"cos(n), here I\u0027ve written the arc cosine the other way."},{"Start":"02:37.190 ","End":"02:44.170","Text":"Similarly with the m, we can make a substitution where we let this equal Theta."},{"Start":"02:44.170 ","End":"02:47.138","Text":"Similarly, this is Theta,"},{"Start":"02:47.138 ","End":"02:51.095","Text":"and it turns out that d Theta is,"},{"Start":"02:51.095 ","End":"02:53.210","Text":"well, not quite what\u0027s here,"},{"Start":"02:53.210 ","End":"02:54.860","Text":"but with a minus."},{"Start":"02:54.860 ","End":"02:57.860","Text":"You might say, \"Where did the minus go?\""},{"Start":"02:57.860 ","End":"03:00.920","Text":"Well, when we substitute the limits,"},{"Start":"03:00.920 ","End":"03:05.255","Text":"we get the limits from Pi to 0."},{"Start":"03:05.255 ","End":"03:08.495","Text":"So instead of from Pi to 0 with a minus,"},{"Start":"03:08.495 ","End":"03:11.840","Text":"we invert, make it from 0 to Pi,"},{"Start":"03:11.840 ","End":"03:13.490","Text":"and then we don\u0027t need the minus,"},{"Start":"03:13.490 ","End":"03:15.185","Text":"so we have this integral."},{"Start":"03:15.185 ","End":"03:17.240","Text":"Now in the previous exercise,"},{"Start":"03:17.240 ","End":"03:21.607","Text":"we computed the integral from minus Pi to Pi of this,"},{"Start":"03:21.607 ","End":"03:23.600","Text":"but since it\u0027s an even function,"},{"Start":"03:23.600 ","End":"03:26.420","Text":"it\u0027s half the integral from minus Pi to Pi."},{"Start":"03:26.420 ","End":"03:28.850","Text":"Now I\u0027ve jumped to that previous exercise,"},{"Start":"03:28.850 ","End":"03:34.220","Text":"but you can see that we showed that the integral of sine could be"},{"Start":"03:34.220 ","End":"03:40.175","Text":"nx and mx is Pi if j equals k, and 0 otherwise."},{"Start":"03:40.175 ","End":"03:46.230","Text":"Now in our case, n is not equal to m and so this just comes out to be 0,"},{"Start":"03:46.230 ","End":"03:48.385","Text":"or a half of 0."},{"Start":"03:48.385 ","End":"03:51.320","Text":"Now we\u0027ll take the case where it\u0027s the same"},{"Start":"03:51.320 ","End":"03:56.893","Text":"n. Instead of the integral of cos(n Theta) cos(m Theta),"},{"Start":"03:56.893 ","End":"04:00.500","Text":"this time we have cos(n Theta)^2."},{"Start":"04:00.500 ","End":"04:03.550","Text":"As we saw, this integral comes out to be Pi,"},{"Start":"04:03.550 ","End":"04:05.680","Text":"so the answer is Pi over 2."},{"Start":"04:05.680 ","End":"04:11.945","Text":"What we want is to find Alpha n such that the norm of Alpha n T_n is 1."},{"Start":"04:11.945 ","End":"04:13.620","Text":"If this norm is 1,"},{"Start":"04:13.620 ","End":"04:16.030","Text":"then the norm squared is 1 and the norm squared"},{"Start":"04:16.030 ","End":"04:18.880","Text":"is the inner product of this with itself,"},{"Start":"04:18.880 ","End":"04:20.470","Text":"so we get this."},{"Start":"04:20.470 ","End":"04:23.815","Text":"Then we can pull out Alpha n squared."},{"Start":"04:23.815 ","End":"04:26.260","Text":"Normally, this scalar comes out"},{"Start":"04:26.260 ","End":"04:29.750","Text":"conjugated but it\u0027s going to be real so it doesn\u0027t matter,"},{"Start":"04:29.750 ","End":"04:31.816","Text":"so it\u0027s just Alpha n squared."},{"Start":"04:31.816 ","End":"04:36.650","Text":"We already have T_n with T_n is Pi over 2,"},{"Start":"04:36.650 ","End":"04:41.220","Text":"so we get Alpha n squared times Pi over 2 equals 1,"},{"Start":"04:41.220 ","End":"04:46.615","Text":"which means that Alpha n squared Pi over 2 is 1."},{"Start":"04:46.615 ","End":"04:52.130","Text":"Which means that Alpha n is the square root of 2 over pi,"},{"Start":"04:52.130 ","End":"04:57.930","Text":"and this is true for all n. That concludes this exercise."}],"ID":28704},{"Watched":false,"Name":"Exercise 8a - Hermite Polynomials","Duration":"6m 58s","ChapterTopicVideoID":27471,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.610","Text":"In this exercise, K is a space of piecewise continuous functions on"},{"Start":"00:05.610 ","End":"00:11.040","Text":"the reals and complex-valued which satisfy this condition."},{"Start":"00:11.040 ","End":"00:13.110","Text":"I\u0027ll return to this in a moment."},{"Start":"00:13.110 ","End":"00:20.415","Text":"The inner product is given by f product with g is the integral of f,"},{"Start":"00:20.415 ","End":"00:24.315","Text":"g conjugate times this function,"},{"Start":"00:24.315 ","End":"00:25.890","Text":"and when we put something here,"},{"Start":"00:25.890 ","End":"00:27.899","Text":"it\u0027s called the weight function."},{"Start":"00:27.899 ","End":"00:37.235","Text":"Returning here, what this condition says is that the norm of f is finite."},{"Start":"00:37.235 ","End":"00:39.455","Text":"Well, this is the norm squared,"},{"Start":"00:39.455 ","End":"00:42.559","Text":"it is the same thing to say that norm squared is finite."},{"Start":"00:42.559 ","End":"00:44.880","Text":"Because if you put f and f here,"},{"Start":"00:44.880 ","End":"00:48.454","Text":"f bar is absolute value of f ^2."},{"Start":"00:48.454 ","End":"00:51.105","Text":"We define for each n,"},{"Start":"00:51.105 ","End":"00:56.805","Text":"H_ n(x), H for Hermite is minus 1^n,"},{"Start":"00:56.805 ","End":"01:00.575","Text":"e^x^2 times the nth derivative of each to the minus x^2,"},{"Start":"01:00.575 ","End":"01:06.140","Text":"and our task is to prove that the set of Hermite polynomials,"},{"Start":"01:06.140 ","End":"01:10.802","Text":"where it goes to 0 to infinity is an orthogonal system in K,"},{"Start":"01:10.802 ","End":"01:14.210","Text":"and then we have to find the normalizing constants for them."},{"Start":"01:14.210 ","End":"01:18.620","Text":"In other words, the constants that if I put a constant Alpha n in here,"},{"Start":"01:18.620 ","End":"01:22.765","Text":"Alpha n, H_ n will be an orthonormal set."},{"Start":"01:22.765 ","End":"01:27.460","Text":"Now, each H_ n is a degree n polynomial."},{"Start":"01:27.460 ","End":"01:29.060","Text":"Won\u0027t prove it formally,"},{"Start":"01:29.060 ","End":"01:30.740","Text":"I\u0027ll give you a hand-waving proof,"},{"Start":"01:30.740 ","End":"01:37.320","Text":"and we differentiate this ones we get minus 2x,"},{"Start":"01:37.510 ","End":"01:43.910","Text":"e^-x^2, if we multiply it by e^x^2,"},{"Start":"01:43.910 ","End":"01:45.200","Text":"you just get minus 2x,"},{"Start":"01:45.200 ","End":"01:47.000","Text":"which is a polynomial of degree 1,"},{"Start":"01:47.000 ","End":"01:51.080","Text":"and then if you differentiate again using the product rule and the chain rule,"},{"Start":"01:51.080 ","End":"01:54.905","Text":"you get a degree 2 polynomial times e^-x^2."},{"Start":"01:54.905 ","End":"01:59.905","Text":"Finally, after n times get a degree n polynomial e^-x^2,"},{"Start":"01:59.905 ","End":"02:03.410","Text":"and then the e^-x^2 cancels with e^x^2,"},{"Start":"02:03.410 ","End":"02:07.265","Text":"leaving just a degree n polynomial and the -1^n."},{"Start":"02:07.265 ","End":"02:09.845","Text":"That\u0027s just an informal proof anyway,"},{"Start":"02:09.845 ","End":"02:11.540","Text":"it\u0027s a degree n polynomial."},{"Start":"02:11.540 ","End":"02:16.490","Text":"What we want to show is that the set is orthogonal,"},{"Start":"02:16.490 ","End":"02:23.600","Text":"in other words that H_ n inner product with H_ m is 0 provided n is not equal to m. Now,"},{"Start":"02:23.600 ","End":"02:25.715","Text":"by a previous exercise,"},{"Start":"02:25.715 ","End":"02:29.660","Text":"it\u0027s sufficient to show that the inner product of H_ n with"},{"Start":"02:29.660 ","End":"02:36.040","Text":"each monomial x^k is 0 whenever k is less than n,"},{"Start":"02:36.040 ","End":"02:39.500","Text":"and in case you don\u0027t remember this previous exercise,"},{"Start":"02:39.500 ","End":"02:40.985","Text":"I\u0027ll remind you,"},{"Start":"02:40.985 ","End":"02:44.270","Text":"and this is the exercise with solution."},{"Start":"02:44.270 ","End":"02:46.730","Text":"Let\u0027s return, we showed this."},{"Start":"02:46.730 ","End":"02:48.650","Text":"This is bit easier to show."},{"Start":"02:48.650 ","End":"02:50.360","Text":"We just have to show this now."},{"Start":"02:50.360 ","End":"02:53.119","Text":"Let\u0027s start with k equals 0,"},{"Start":"02:53.119 ","End":"02:57.200","Text":"H_ n, with x^ 0 is 1."},{"Start":"02:57.200 ","End":"03:00.875","Text":"This formula, just using the inner product formula,"},{"Start":"03:00.875 ","End":"03:09.240","Text":"it\u0027s product of H_ n(x) times 1 times the weight function and the integral,"},{"Start":"03:09.240 ","End":"03:12.799","Text":"and this is equal to H_ n, which I\u0027ve colored,"},{"Start":"03:12.799 ","End":"03:16.345","Text":"is equal to this colored bit,"},{"Start":"03:16.345 ","End":"03:18.615","Text":"and then stuff cancels,"},{"Start":"03:18.615 ","End":"03:22.785","Text":"e^x squared cancels with e^ minus x squared,"},{"Start":"03:22.785 ","End":"03:26.355","Text":"minus 1^n comes in front of the integral,"},{"Start":"03:26.355 ","End":"03:28.160","Text":"and we have this expression,"},{"Start":"03:28.160 ","End":"03:33.110","Text":"the nth derivative of e^minus x squared here in the integrand."},{"Start":"03:33.110 ","End":"03:39.530","Text":"Now, the indefinite integral of this is just the n minus 1th,"},{"Start":"03:39.530 ","End":"03:42.350","Text":"n minus first derivative."},{"Start":"03:42.350 ","End":"03:45.014","Text":"Differentiate n times and integrate once,"},{"Start":"03:45.014 ","End":"03:48.460","Text":"you\u0027re left with the n minus 1 derivative,"},{"Start":"03:48.460 ","End":"03:50.840","Text":"and so now instead of the integral,"},{"Start":"03:50.840 ","End":"03:55.910","Text":"we just have to substitute the limits minus infinity to infinity."},{"Start":"03:55.910 ","End":"03:59.630","Text":"Doesn\u0027t seem so simple to figure this out,"},{"Start":"03:59.630 ","End":"04:00.740","Text":"but here\u0027s how we do."},{"Start":"04:00.740 ","End":"04:04.115","Text":"It. Turns out, I will show this,"},{"Start":"04:04.115 ","End":"04:07.700","Text":"that the n minus 1th derivative of e^-x^2"},{"Start":"04:07.700 ","End":"04:14.720","Text":"is some polynomial of x times e^-x^2."},{"Start":"04:14.720 ","End":"04:17.690","Text":"We\u0027ll prove this claim and then we\u0027ll return here."},{"Start":"04:17.690 ","End":"04:21.470","Text":"This claim will prove by induction on n. First of all,"},{"Start":"04:21.470 ","End":"04:24.005","Text":"take the case n equals 1,"},{"Start":"04:24.005 ","End":"04:26.420","Text":"and in that case n minus 1 is 0."},{"Start":"04:26.420 ","End":"04:29.360","Text":"We got the 0th derivative of e^-x^2,"},{"Start":"04:29.360 ","End":"04:31.700","Text":"which is just e^-x^2,"},{"Start":"04:31.700 ","End":"04:34.640","Text":"and we could let p(x) be the constant polynomial 1,"},{"Start":"04:34.640 ","End":"04:37.055","Text":"so that\u0027s okay for n equals 1."},{"Start":"04:37.055 ","End":"04:40.565","Text":"Now we need the induction step from n to n plus 1."},{"Start":"04:40.565 ","End":"04:43.715","Text":"If we replace n by n plus 1 here,"},{"Start":"04:43.715 ","End":"04:51.230","Text":"we have the nth derivative and this equals d by dx of the n minus 1th derivative."},{"Start":"04:51.230 ","End":"04:53.855","Text":"By the induction hypothesis,"},{"Start":"04:53.855 ","End":"04:59.630","Text":"this is equal to what\u0027s inside the bracket as p(x)e^-x^2,"},{"Start":"04:59.630 ","End":"05:01.580","Text":"where p(x) is some polynomial,"},{"Start":"05:01.580 ","End":"05:06.605","Text":"and if we differentiate this, we get p\u0027(x)."},{"Start":"05:06.605 ","End":"05:10.910","Text":"It\u0027s a product, but it\u0027s also a chain rule."},{"Start":"05:10.910 ","End":"05:13.790","Text":"Then p(x) not differentiated,"},{"Start":"05:13.790 ","End":"05:15.200","Text":"and this one differentiated,"},{"Start":"05:15.200 ","End":"05:18.110","Text":"which gives us -2x, e-x^2,"},{"Start":"05:18.110 ","End":"05:21.080","Text":"and the minus e^ x^2 is collected outside the brackets,"},{"Start":"05:21.080 ","End":"05:26.540","Text":"and this is also a polynomial, call it P(x)."},{"Start":"05:26.540 ","End":"05:30.740","Text":"Indeed, this nth derivative of e^ minus x"},{"Start":"05:30.740 ","End":"05:34.835","Text":"squared is a polynomial times e^ minus x squared also,"},{"Start":"05:34.835 ","End":"05:37.715","Text":"QED for this claim."},{"Start":"05:37.715 ","End":"05:40.010","Text":"Now let\u0027s get back here."},{"Start":"05:40.010 ","End":"05:44.915","Text":"This was H_ n inner product with 1."},{"Start":"05:44.915 ","End":"05:46.595","Text":"This is equal to,"},{"Start":"05:46.595 ","End":"05:48.444","Text":"just copying this,"},{"Start":"05:48.444 ","End":"05:52.575","Text":"and now we want to apply this claim that we just proved."},{"Start":"05:52.575 ","End":"05:56.055","Text":"We have minus 1^n here,"},{"Start":"05:56.055 ","End":"06:01.245","Text":"and then this n minus 1th derivative is P(x)e^x^2,"},{"Start":"06:01.245 ","End":"06:05.885","Text":"and we have to take this between the limits of minus infinity to infinity."},{"Start":"06:05.885 ","End":"06:12.950","Text":"Now, I claim that both infinity and minus infinity when substituted here give us 0."},{"Start":"06:12.950 ","End":"06:15.110","Text":"When I say substitute, I mean that loosely,"},{"Start":"06:15.110 ","End":"06:18.949","Text":"I mean next turn to infinity or minus infinity."},{"Start":"06:18.949 ","End":"06:24.065","Text":"But either way, P(x)e^-x^ goes to 0"},{"Start":"06:24.065 ","End":"06:30.015","Text":"because it\u0027s like P(x) in the numerator and e^x^2 in the denominator,"},{"Start":"06:30.015 ","End":"06:37.085","Text":"and an exponent goes to infinity faster than a polynomial however large the degree here,"},{"Start":"06:37.085 ","End":"06:39.185","Text":"and so this turns to 0."},{"Start":"06:39.185 ","End":"06:42.905","Text":"Actually we get 0 minus 0, which is 0."},{"Start":"06:42.905 ","End":"06:44.630","Text":"To put things in perspective,"},{"Start":"06:44.630 ","End":"06:52.675","Text":"what we\u0027ve just proved is the case where K equals 0 of this claim."},{"Start":"06:52.675 ","End":"06:58.900","Text":"So next we\u0027re going to return to this claim where K is bigger than 0."}],"ID":28705},{"Watched":false,"Name":"Exercise 8b - Hermite Polynomials","Duration":"9m 47s","ChapterTopicVideoID":27472,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.430","Text":"Continuing the previous clip,"},{"Start":"00:02.430 ","End":"00:05.310","Text":"we now turn to the case k is bigger than 0."},{"Start":"00:05.310 ","End":"00:07.755","Text":"We just did the case k=0."},{"Start":"00:07.755 ","End":"00:12.675","Text":"The inner product of H_n with x^k is this integral."},{"Start":"00:12.675 ","End":"00:16.245","Text":"Replace it with what it\u0027s equal to in red here."},{"Start":"00:16.245 ","End":"00:19.380","Text":"Now stuff cancels e^(x^2), with e^(minus x^2)."},{"Start":"00:19.380 ","End":"00:25.770","Text":"Minus 1^n will bring in front of the integral so"},{"Start":"00:25.770 ","End":"00:32.250","Text":"we end up with this and going do this integration by parts is the formula to remind you,"},{"Start":"00:32.250 ","End":"00:37.125","Text":"we\u0027re going to let this be f\u0027 and this g. Using this formula,"},{"Start":"00:37.125 ","End":"00:44.575","Text":"we then get f(g) evaluated between the limits minus the integral of f(g\u0027)."},{"Start":"00:44.575 ","End":"00:49.610","Text":"Maybe the only thing to comment on is that the integral of"},{"Start":"00:49.610 ","End":"00:56.360","Text":"the nth derivative is just the n minus 1 for n minus first derivative."},{"Start":"00:56.360 ","End":"01:00.990","Text":"Now, I\u0027m going to leave the developing of this equality and just"},{"Start":"01:00.990 ","End":"01:07.010","Text":"focus on the part I have marked in blue, this part here."},{"Start":"01:07.010 ","End":"01:14.420","Text":"Now we said earlier that this n minus 1\u0027s derivative is equal to some polynomial in"},{"Start":"01:14.420 ","End":"01:22.395","Text":"x times e^(minus x^2) and we have an extra x^k here but it\u0027s still a polynomial."},{"Start":"01:22.395 ","End":"01:25.735","Text":"P(x) times x^k is another polynomial, q(x)."},{"Start":"01:25.735 ","End":"01:27.230","Text":"So we have a polynomial in"},{"Start":"01:27.230 ","End":"01:32.780","Text":"x times e^(minus x^2) evaluated from minus infinity to infinity."},{"Start":"01:32.780 ","End":"01:38.975","Text":"This is equal to 0 because if you substitute infinity or minus infinity you get 0."},{"Start":"01:38.975 ","End":"01:45.650","Text":"Because what we have is q(x) divided by e^(x^2) polynomial over exponential."},{"Start":"01:45.650 ","End":"01:49.790","Text":"The exponential wins and when it goes to plus or minus infinity it goes to 0,"},{"Start":"01:49.790 ","End":"01:51.290","Text":"so this part is 0."},{"Start":"01:51.290 ","End":"01:53.290","Text":"Now we can go back up here,"},{"Start":"01:53.290 ","End":"02:02.240","Text":"we have this was H_n^x^k is equal to,"},{"Start":"02:02.240 ","End":"02:06.560","Text":"this part disappears and we just get the second part."},{"Start":"02:06.560 ","End":"02:09.500","Text":"Well, I\u0027ll just put a 0 if we placeholder for this,"},{"Start":"02:09.500 ","End":"02:13.220","Text":"then copy this second bit with the minus."},{"Start":"02:13.220 ","End":"02:15.680","Text":"Do a little bit of simplifying."},{"Start":"02:15.680 ","End":"02:23.340","Text":"The minus here with the minus 1^n is minus 1^n minus 1."},{"Start":"02:23.480 ","End":"02:27.535","Text":"The k here goes in front of the integral,"},{"Start":"02:27.535 ","End":"02:30.220","Text":"so now we have this expression,"},{"Start":"02:30.220 ","End":"02:37.360","Text":"the n minus 1\u0027s derivative of e^(minus x^2), x^(k minus 1)."},{"Start":"02:37.360 ","End":"02:39.871","Text":"Now I\u0027m going to rearrange it a bit. You know what?"},{"Start":"02:39.871 ","End":"02:42.370","Text":"I\u0027ll do the following step and you\u0027ll see what I\u0027m doing."},{"Start":"02:42.370 ","End":"02:46.675","Text":"I\u0027m working backwards to get H_n minus 1."},{"Start":"02:46.675 ","End":"02:52.060","Text":"H_n minus 1(x) is this expression that\u0027s in red."},{"Start":"02:52.060 ","End":"02:59.440","Text":"Just put the minus 1^(n minus 1) back in and we needed an e^(x^2) here,"},{"Start":"02:59.440 ","End":"03:03.625","Text":"compensated by e^( minus x^2) here."},{"Start":"03:03.625 ","End":"03:05.320","Text":"Everything else has this."},{"Start":"03:05.320 ","End":"03:08.610","Text":"What that gives us is k times the inner product of H_n"},{"Start":"03:08.610 ","End":"03:12.270","Text":"minus 1 with the monomial x^(k minus 1)."},{"Start":"03:12.270 ","End":"03:15.135","Text":"This is a recursive relation."},{"Start":"03:15.135 ","End":"03:19.320","Text":"We have this relation now our aim is to show that this is"},{"Start":"03:19.320 ","End":"03:24.380","Text":"0 and we\u0027ll work by induction or recursion and assume that this is 0."},{"Start":"03:24.380 ","End":"03:28.430","Text":"What we\u0027re doing is we\u0027re doing it by induction on n and we\u0027ve"},{"Start":"03:28.430 ","End":"03:32.600","Text":"already taken care of the case n=1."},{"Start":"03:32.600 ","End":"03:37.580","Text":"If n=1, then k=0 and we proved the case k=0."},{"Start":"03:37.580 ","End":"03:39.710","Text":"We\u0027ve proved the k, n=1."},{"Start":"03:39.710 ","End":"03:45.005","Text":"Now also note here that k is less than n"},{"Start":"03:45.005 ","End":"03:51.170","Text":"so k minus 1 is less than n minus 1 so we really do have a recursion situation."},{"Start":"03:51.170 ","End":"03:55.870","Text":"Anyway, using the induction on the case n minus 1,"},{"Start":"03:55.870 ","End":"04:00.680","Text":"we have that H_n minus 1x^(k minus 1) inner product is 0."},{"Start":"04:00.680 ","End":"04:06.340","Text":"Plug that in here and we\u0027ve got that this is k times 0, which is 0."},{"Start":"04:06.340 ","End":"04:11.120","Text":"This is what we had to show in order to show that the H_ns are orthogonal."},{"Start":"04:11.120 ","End":"04:15.390","Text":"Remember we reduced it to this using a previous exercise."},{"Start":"04:15.390 ","End":"04:20.675","Text":"Now we\u0027ve shown that the set H_n and goes from 0 to infinity,"},{"Start":"04:20.675 ","End":"04:22.715","Text":"it\u0027s an orthogonal set."},{"Start":"04:22.715 ","End":"04:27.030","Text":"Now we\u0027ll find the normalizing constants."},{"Start":"04:27.030 ","End":"04:31.710","Text":"The concepts we\u0027re looking for are what we multiply the H_n by to get"},{"Start":"04:31.710 ","End":"04:38.125","Text":"an orthonormal system so we want the norm of Alpha n H_n to be 1."},{"Start":"04:38.125 ","End":"04:41.405","Text":"Now, if we take the inner product of H_n with H_n,"},{"Start":"04:41.405 ","End":"04:43.430","Text":"one of the H_n is the one on the right."},{"Start":"04:43.430 ","End":"04:47.285","Text":"I can replace with the polynomial equivalent,"},{"Start":"04:47.285 ","End":"04:51.520","Text":"some polynomial of degree n. We can expand"},{"Start":"04:51.520 ","End":"04:56.195","Text":"this linearly but we have to bring the a_n out with the conjugate."},{"Start":"04:56.195 ","End":"05:00.050","Text":"Now, only the first one matters because we\u0027ve"},{"Start":"05:00.050 ","End":"05:03.665","Text":"shown that the inner product of H_n with x^k,"},{"Start":"05:03.665 ","End":"05:07.600","Text":"where k is less and n is 0 so all these are 0."},{"Start":"05:07.600 ","End":"05:12.205","Text":"What we\u0027re left with here is just a_n bar H_n, x^n."},{"Start":"05:12.205 ","End":"05:20.090","Text":"Now I\u0027m going to show you by induction that this coefficient a_n is 2^n."},{"Start":"05:20.090 ","End":"05:24.755","Text":"If n=0, we get that H_0 is just equal to 1,"},{"Start":"05:24.755 ","End":"05:33.020","Text":"which is 2^0 and this is equal to a_n or a_0 because if H_n is the polynomial 1,"},{"Start":"05:33.020 ","End":"05:37.855","Text":"then all of this is just a_0 which is 1."},{"Start":"05:37.855 ","End":"05:42.875","Text":"That\u0027s the case 0. Now, when n is bigger than 0,"},{"Start":"05:42.875 ","End":"05:46.455","Text":"we have the following H_n is equal to this."},{"Start":"05:46.455 ","End":"05:52.945","Text":"We multiply both sides by e^( minus x^2) so this e^(x^2) disappears."},{"Start":"05:52.945 ","End":"05:56.560","Text":"Now differentiate both sides."},{"Start":"05:56.560 ","End":"05:58.200","Text":"This is a product,"},{"Start":"05:58.200 ","End":"06:01.270","Text":"so it\u0027s H_n\u0027 e^(minus x^2),"},{"Start":"06:02.210 ","End":"06:07.785","Text":"and then minus 2x e^( minus x^2) times H_n."},{"Start":"06:07.785 ","End":"06:10.305","Text":"Again, this comes out."},{"Start":"06:10.305 ","End":"06:15.120","Text":"Here we just increase by 1 the derivative."},{"Start":"06:15.120 ","End":"06:20.835","Text":"Now bring the e^( minus x^2) again back to the other side as e^(x^2)."},{"Start":"06:20.835 ","End":"06:27.230","Text":"Then the right-hand expression is exactly the definition of H_n plus 1 so we have"},{"Start":"06:27.230 ","End":"06:30.935","Text":"this relation which is a recursion relation"},{"Start":"06:30.935 ","End":"06:36.585","Text":"that gives H_n plus 1 in terms of H_n and H_n\u0027."},{"Start":"06:36.585 ","End":"06:42.885","Text":"Now, H_n we said is some polynomial we call the coefficients a_n,"},{"Start":"06:42.885 ","End":"06:45.585","Text":"a_n x^n down to a_0."},{"Start":"06:45.585 ","End":"06:48.810","Text":"I\u0027m just plugging it in here and here,"},{"Start":"06:48.810 ","End":"06:55.710","Text":"the derivative na_n^x-1 down to a_1."},{"Start":"06:55.710 ","End":"07:01.100","Text":"H_n plus 1 is an n plus 1-degree polynomial called the coefficients b_n."},{"Start":"07:01.100 ","End":"07:03.890","Text":"It goes up to b_n plus 1x^n plus 1."},{"Start":"07:03.890 ","End":"07:05.900","Text":"Now compare the highest coefficient,"},{"Start":"07:05.900 ","End":"07:08.695","Text":"which is the coefficient of x^n plus 1."},{"Start":"07:08.695 ","End":"07:14.235","Text":"From this side, we get 2a_n x^n plus 1."},{"Start":"07:14.235 ","End":"07:17.355","Text":"Here we have b^n plus 1 x^n plus 1."},{"Start":"07:17.355 ","End":"07:21.380","Text":"We can compare coefficients and by"},{"Start":"07:21.380 ","End":"07:29.435","Text":"the induction hypothesis a_n is 2^n so b_n plus 1 is 2^(n plus 1)."},{"Start":"07:29.435 ","End":"07:32.330","Text":"Which means that the leading coefficient for the case n"},{"Start":"07:32.330 ","End":"07:35.509","Text":"plus 1 is 2^(n plus 1) as required."},{"Start":"07:35.509 ","End":"07:37.970","Text":"Now it\u0027s not the letter H it\u0027s different letter b,"},{"Start":"07:37.970 ","End":"07:42.050","Text":"but what we\u0027re really saying is that the highest coefficient is 2^(n plus 1),"},{"Start":"07:42.050 ","End":"07:43.520","Text":"it doesn\u0027t matter what letter."},{"Start":"07:43.520 ","End":"07:45.905","Text":"Now look back a minute."},{"Start":"07:45.905 ","End":"07:49.295","Text":"This is what I want. H_n,"},{"Start":"07:49.295 ","End":"07:55.490","Text":"H_n is equal to a_n bar H_n, x^n."},{"Start":"07:55.490 ","End":"07:56.945","Text":"If we use that,"},{"Start":"07:56.945 ","End":"08:01.280","Text":"we get replacing a_n bar with 2^n,"},{"Start":"08:01.280 ","End":"08:03.199","Text":"we get this relation."},{"Start":"08:03.199 ","End":"08:09.557","Text":"Continuing. Next thing we\u0027ll do is show by induction that this H_n,"},{"Start":"08:09.557 ","End":"08:13.730","Text":"x^n is equal to n factorial square root of Pi."},{"Start":"08:13.730 ","End":"08:16.990","Text":"As always by induction, for n=0,"},{"Start":"08:16.990 ","End":"08:19.620","Text":"we get a product of 1 with 1,"},{"Start":"08:19.620 ","End":"08:21.150","Text":"which is this integral."},{"Start":"08:21.150 ","End":"08:24.540","Text":"We mentioned in the beginning that this integral is square root of"},{"Start":"08:24.540 ","End":"08:28.665","Text":"Pi so it fits n factorial square root of Pi,"},{"Start":"08:28.665 ","End":"08:30.855","Text":"0 factorial root Pi,"},{"Start":"08:30.855 ","End":"08:32.635","Text":"0 factorial is 1."},{"Start":"08:32.635 ","End":"08:35.120","Text":"Now let\u0027s take the case where n is bigger than 0."},{"Start":"08:35.120 ","End":"08:40.715","Text":"We showed earlier that we have this recurrence relation and if we have that,"},{"Start":"08:40.715 ","End":"08:44.105","Text":"then in particular letting k=n,"},{"Start":"08:44.105 ","End":"08:45.127","Text":"we have H_n,"},{"Start":"08:45.127 ","End":"08:49.730","Text":"x^n is n(H_(n minus 1), x^(n minus 1))."},{"Start":"08:49.730 ","End":"08:53.419","Text":"This is equal to n times by the induction,"},{"Start":"08:53.419 ","End":"08:55.660","Text":"this is equal to n minus 1 factorial,"},{"Start":"08:55.660 ","End":"08:57.485","Text":"square root of Pi,"},{"Start":"08:57.485 ","End":"09:02.120","Text":"and n times n minus 1 factorial is n factorial."},{"Start":"09:02.120 ","End":"09:04.414","Text":"This is the inductive step."},{"Start":"09:04.414 ","End":"09:06.650","Text":"Now, a reminder of what we had a short while ago,"},{"Start":"09:06.650 ","End":"09:09.300","Text":"we had this relation that H_n,"},{"Start":"09:09.300 ","End":"09:11.695","Text":"H_n is 2(H_n, x^n)."},{"Start":"09:11.695 ","End":"09:13.727","Text":"Now, this is equal to H_n,"},{"Start":"09:13.727 ","End":"09:17.155","Text":"x^n from here is n factorial square root of Pi."},{"Start":"09:17.155 ","End":"09:20.460","Text":"We have the norm of (H_n)^2 is this,"},{"Start":"09:20.460 ","End":"09:24.995","Text":"and that means that the norm of H_n is the square root of this."},{"Start":"09:24.995 ","End":"09:30.050","Text":"Recall that we want the norm of Alpha and H_n to be 1 and we can take"},{"Start":"09:30.050 ","End":"09:33.230","Text":"the Alpha n outside the norm because it\u0027s"},{"Start":"09:33.230 ","End":"09:38.030","Text":"a positive number and then we can divide by the norm of H_n."},{"Start":"09:38.030 ","End":"09:44.890","Text":"What we get finally is that Alpha n is 1 over this expression."},{"Start":"09:44.890 ","End":"09:48.570","Text":"We\u0027re finally done with this exercise."}],"ID":28706},{"Watched":false,"Name":"Exercise 9a - Legendre Polynomials","Duration":"8m 4s","ChapterTopicVideoID":27473,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.240","Text":"In this exercise, K is the space of"},{"Start":"00:03.240 ","End":"00:08.665","Text":"piecewise continuous functions on the open interval minus 1, 1."},{"Start":"00:08.665 ","End":"00:15.330","Text":"The inner product is the usual f times g conjugate integral on the interval."},{"Start":"00:15.330 ","End":"00:17.670","Text":"There\u0027s no weight function this time,"},{"Start":"00:17.670 ","End":"00:19.920","Text":"or you could say the weight function is 1."},{"Start":"00:19.920 ","End":"00:25.095","Text":"We define polynomials P_n(x) by the following formula,"},{"Start":"00:25.095 ","End":"00:31.365","Text":"the nth derivative of x^2 minus 1^n times 1 over 2^n, n factorial."},{"Start":"00:31.365 ","End":"00:34.815","Text":"These are called Legendre polynomials."},{"Start":"00:34.815 ","End":"00:37.460","Text":"What we have to do is to prove that this set,"},{"Start":"00:37.460 ","End":"00:40.350","Text":"P_n from 0 to infinity,"},{"Start":"00:40.350 ","End":"00:44.730","Text":"is an orthogonal system in K and also that the norm"},{"Start":"00:44.730 ","End":"00:51.075","Text":"of P_n^2 in a product with P_n is 2 over 2n plus 1."},{"Start":"00:51.075 ","End":"00:55.370","Text":"First, let me note that P_n really is a degree n polynomial"},{"Start":"00:55.370 ","End":"01:00.379","Text":"because it\u0027s the nth derivative of a degree 2n polynomial."},{"Start":"01:00.379 ","End":"01:03.085","Text":"This is x^2 to the power of n, 2n."},{"Start":"01:03.085 ","End":"01:05.240","Text":"Just for interest sake,"},{"Start":"01:05.240 ","End":"01:09.150","Text":"here are the first few Legendre polynomials."},{"Start":"01:09.150 ","End":"01:16.610","Text":"What we have to show is that the inner product of P_n with P_m is 0 if n is not equal to"},{"Start":"01:16.610 ","End":"01:24.112","Text":"m. It was a previous exercise where we showed that it\u0027s enough for this to be true,"},{"Start":"01:24.112 ","End":"01:27.220","Text":"that the inner product of P_n with x^k,"},{"Start":"01:27.220 ","End":"01:36.430","Text":"the monomial is 0 as long as k is less than n. Here is this exercise just for reference,"},{"Start":"01:36.430 ","End":"01:39.610","Text":"the statement, and the proof."},{"Start":"01:39.610 ","End":"01:42.610","Text":"Let\u0027s get back to our exercise."},{"Start":"01:42.610 ","End":"01:44.364","Text":"Let\u0027s start computing."},{"Start":"01:44.364 ","End":"01:51.470","Text":"The inner product of P_n with x^k is the integral of P_n(x) times x^k."},{"Start":"01:51.470 ","End":"01:54.760","Text":"This is equal to, by the definition,"},{"Start":"01:54.760 ","End":"01:58.335","Text":"this, just reverse the order of these 2."},{"Start":"01:58.335 ","End":"02:01.734","Text":"Now we\u0027re going to use integration by parts."},{"Start":"02:01.734 ","End":"02:05.040","Text":"This will be f’ and this will be g,"},{"Start":"02:05.040 ","End":"02:08.550","Text":"so what we get is f,"},{"Start":"02:08.550 ","End":"02:17.250","Text":"which is just the n minus 1th derivative and g is here and that evaluated from minus 1 to"},{"Start":"02:17.250 ","End":"02:21.015","Text":"1 minus f again"},{"Start":"02:21.015 ","End":"02:28.535","Text":"g’ which is k x^k minus 1 and put the k in front of the integral."},{"Start":"02:28.535 ","End":"02:32.420","Text":"Now, note that this bit that I\u0027ve highlighted,"},{"Start":"02:32.420 ","End":"02:39.865","Text":"the claim is that this is equal to some polynomial of x times x^2 minus 1."},{"Start":"02:39.865 ","End":"02:43.790","Text":"The way we\u0027ll prove this is by induction."},{"Start":"02:43.790 ","End":"02:47.165","Text":"Instead of the n minus 1 here and here,"},{"Start":"02:47.165 ","End":"02:49.120","Text":"we\u0027ll put k,"},{"Start":"02:49.120 ","End":"02:56.390","Text":"and we\u0027ll prove it by induction for k between 0 and n. Afterward,"},{"Start":"02:56.390 ","End":"02:59.573","Text":"we\u0027ll substitute k equals n minus 1,"},{"Start":"02:59.573 ","End":"03:02.970","Text":"and so we\u0027ll get the n minus 1 here."},{"Start":"03:02.970 ","End":"03:05.340","Text":"When k is n minus 1,"},{"Start":"03:05.340 ","End":"03:06.735","Text":"we\u0027ll get a 1 here."},{"Start":"03:06.735 ","End":"03:09.705","Text":"Well, we haven\u0027t written the 1 but it\u0027s there."},{"Start":"03:09.705 ","End":"03:14.415","Text":"Instead of p(x) would have p_k(x),"},{"Start":"03:14.415 ","End":"03:20.735","Text":"where the degree of p_k is k. That\u0027s the base case of the induction."},{"Start":"03:20.735 ","End":"03:23.930","Text":"Now the induction step from k to k plus 1,"},{"Start":"03:23.930 ","End":"03:30.300","Text":"but we have to assume that k is still less than n. Replacing k by k plus 1,"},{"Start":"03:30.300 ","End":"03:33.630","Text":"start from the left-hand side and we\u0027ll reach the right-hand side."},{"Start":"03:33.630 ","End":"03:40.475","Text":"The k plus 1th derivative is the derivative of the kth derivative and by the induction,"},{"Start":"03:40.475 ","End":"03:43.220","Text":"the kth derivative is equal to this."},{"Start":"03:43.220 ","End":"03:45.890","Text":"This equal, using the product rule,"},{"Start":"03:45.890 ","End":"03:48.395","Text":"the derivative of this times this,"},{"Start":"03:48.395 ","End":"03:51.715","Text":"plus this times the derivative of this."},{"Start":"03:51.715 ","End":"03:53.820","Text":"We use the chain rule,"},{"Start":"03:53.820 ","End":"03:55.965","Text":"that\u0027s why the 2x is here."},{"Start":"03:55.965 ","End":"03:59.540","Text":"We can take x^2 minus 1 to the power of n minus k"},{"Start":"03:59.540 ","End":"04:04.115","Text":"minus 1 out of the brackets since here it\u0027s one more."},{"Start":"04:04.115 ","End":"04:06.185","Text":"This is what we get."},{"Start":"04:06.185 ","End":"04:10.595","Text":"Now, the degree of the bit in square brackets, the first part,"},{"Start":"04:10.595 ","End":"04:13.730","Text":"it\u0027s k minus 1 plus 2,"},{"Start":"04:13.730 ","End":"04:15.595","Text":"and for the second part,"},{"Start":"04:15.595 ","End":"04:19.490","Text":"it\u0027s 1 plus k. Need the maximum of those."},{"Start":"04:19.490 ","End":"04:20.720","Text":"Well, they\u0027re both k plus 1,"},{"Start":"04:20.720 ","End":"04:24.215","Text":"so the degree here is k plus 1."},{"Start":"04:24.215 ","End":"04:30.590","Text":"This is the polynomial of degree k plus 1 in x times this."},{"Start":"04:30.590 ","End":"04:33.875","Text":"This exactly proves, by induction,"},{"Start":"04:33.875 ","End":"04:38.240","Text":"that this holds because we\u0027ve got here on the right-hand side,"},{"Start":"04:38.240 ","End":"04:42.020","Text":"what we\u0027re supposed to have if you replace k by k plus 1."},{"Start":"04:42.020 ","End":"04:46.220","Text":"In particular, if you replace k by n minus 1,"},{"Start":"04:46.220 ","End":"04:49.490","Text":"then we get what we set out to prove."},{"Start":"04:49.490 ","End":"04:53.300","Text":"Now continuing from before, we had this."},{"Start":"04:53.300 ","End":"04:57.830","Text":"If you go back, you see we got up to this step and then we took a side trip"},{"Start":"04:57.830 ","End":"05:02.600","Text":"to prove that this is equal to this by induction."},{"Start":"05:02.600 ","End":"05:05.150","Text":"Well, except for the x^k bit,"},{"Start":"05:05.150 ","End":"05:06.710","Text":"which we just tack on,"},{"Start":"05:06.710 ","End":"05:11.600","Text":"so what we get is that this bit that\u0027s colored in blue is equal to,"},{"Start":"05:11.600 ","End":"05:13.250","Text":"by the induction step, this,"},{"Start":"05:13.250 ","End":"05:15.200","Text":"but with an x^k here,"},{"Start":"05:15.200 ","End":"05:19.395","Text":"and if we plug in 1 and minus 1, we get 0."},{"Start":"05:19.395 ","End":"05:21.240","Text":"If x is 1 or minus 1 it\u0027s 0,"},{"Start":"05:21.240 ","End":"05:24.000","Text":"so 0 minus 0 which is 0."},{"Start":"05:24.000 ","End":"05:30.890","Text":"So this part in blue just disappears and we have minus k. We can throw that on the 2^n,"},{"Start":"05:30.890 ","End":"05:34.180","Text":"n factorial times this integral."},{"Start":"05:34.180 ","End":"05:37.490","Text":"To summarize, we started with this,"},{"Start":"05:37.490 ","End":"05:43.370","Text":"and then we reached this using integration by parts and other tricks."},{"Start":"05:43.370 ","End":"05:47.425","Text":"Now, if we repeat the process that we did to get from here to here,"},{"Start":"05:47.425 ","End":"05:49.535","Text":"we\u0027ll get the next step,"},{"Start":"05:49.535 ","End":"05:53.450","Text":"which will be instead of minus k,"},{"Start":"05:53.450 ","End":"05:55.880","Text":"we\u0027ll have k times k minus 1."},{"Start":"05:55.880 ","End":"06:00.590","Text":"Each time we multiply by a negative and k minus 1,"},{"Start":"06:00.590 ","End":"06:03.275","Text":"which is taken from the degree here,"},{"Start":"06:03.275 ","End":"06:06.450","Text":"we lower this by 1."},{"Start":"06:06.450 ","End":"06:08.630","Text":"Anyway, we get this after the next step,"},{"Start":"06:08.630 ","End":"06:11.280","Text":"I\u0027ll save you the tedious work."},{"Start":"06:11.280 ","End":"06:13.100","Text":"If we keep going like this,"},{"Start":"06:13.100 ","End":"06:14.345","Text":"instead of the 2,"},{"Start":"06:14.345 ","End":"06:21.950","Text":"we can get k. The good of that is that we get rid of this power of x and we get,"},{"Start":"06:21.950 ","End":"06:23.690","Text":"if we just simplify it a bit,"},{"Start":"06:23.690 ","End":"06:26.165","Text":"this part here is k factorial,"},{"Start":"06:26.165 ","End":"06:30.200","Text":"so this is what we get with no power of x here."},{"Start":"06:30.200 ","End":"06:35.810","Text":"That\u0027s good for us because then we have the integral of a derivative and we can"},{"Start":"06:35.810 ","End":"06:41.210","Text":"use this formula from fundamental theorem of calculus."},{"Start":"06:41.210 ","End":"06:44.455","Text":"We can use it on this part here."},{"Start":"06:44.455 ","End":"06:50.840","Text":"This is our f’, so f itself would be just 1 less."},{"Start":"06:50.840 ","End":"06:54.410","Text":"The same thing as here, but with an n minus k minus 1,"},{"Start":"06:54.410 ","End":"06:57.320","Text":"and we have to evaluate it between minus 1 and 1."},{"Start":"06:57.320 ","End":"07:02.690","Text":"Again, we\u0027ll make use of a previous result we proved in general that we have"},{"Start":"07:02.690 ","End":"07:09.530","Text":"the kth derivative of x^2 minus 1^n which equal to k degree polynomial,"},{"Start":"07:09.530 ","End":"07:15.045","Text":"and x times x^2 minus 1^n minus k. It\u0027s true whenever k is between"},{"Start":"07:15.045 ","End":"07:21.455","Text":"0 and n. If we replace k by n minus k plus 1 which is also between 0 and n,"},{"Start":"07:21.455 ","End":"07:24.175","Text":"then we get the following,"},{"Start":"07:24.175 ","End":"07:27.390","Text":"and this constant if you just repeat it."},{"Start":"07:27.390 ","End":"07:31.970","Text":"So what we\u0027ve shown is that the inner product of P_n with"},{"Start":"07:31.970 ","End":"07:38.506","Text":"x^k is 0 for k less than n. This is what we set out,"},{"Start":"07:38.506 ","End":"07:43.700","Text":"this is the major step because we said that this proves that P_n,"},{"Start":"07:43.700 ","End":"07:47.375","Text":"P_m inner product is 0 for m less than n,"},{"Start":"07:47.375 ","End":"07:55.820","Text":"and hence for m not equal n. What is left to prove is the other bit that P_n,"},{"Start":"07:55.820 ","End":"07:59.660","Text":"P_m in a product is 2 over 2n plus 1."},{"Start":"07:59.660 ","End":"08:05.220","Text":"We\u0027ll take a break and continue proving this after the break."}],"ID":28707},{"Watched":false,"Name":"Exercise 9b - Legendre Polynomials","Duration":"8m 23s","ChapterTopicVideoID":27474,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"We\u0027re back after the break,"},{"Start":"00:01.770 ","End":"00:04.770","Text":"and just to remind you which exercise we\u0027re on,"},{"Start":"00:04.770 ","End":"00:07.980","Text":"you can pause and take a look at it."},{"Start":"00:07.980 ","End":"00:12.420","Text":"What we had to do still is to show that"},{"Start":"00:12.420 ","End":"00:17.980","Text":"the inner product of P_n with P_n is 2 over 2n plus 1."},{"Start":"00:18.080 ","End":"00:24.270","Text":"It\u0027s convenient to take the case n=0 first."},{"Start":"00:24.270 ","End":"00:27.210","Text":"Note that P_naught of x is 1,"},{"Start":"00:27.210 ","End":"00:30.570","Text":"so P_naught product P_naught is 1 with 1,"},{"Start":"00:30.570 ","End":"00:36.855","Text":"which is the integral of 1 times 1dx on the interval from minus 1-1,"},{"Start":"00:36.855 ","End":"00:39.582","Text":"1 minus minus 1 is 2,"},{"Start":"00:39.582 ","End":"00:44.610","Text":"and 2 can be written in this form as 2 over 2 times 0 plus 1."},{"Start":"00:44.610 ","End":"00:47.535","Text":"We\u0027re okay with n=0,"},{"Start":"00:47.535 ","End":"00:51.350","Text":"and now we\u0027ll take the case where n is bigger than 0."},{"Start":"00:51.350 ","End":"00:54.365","Text":"P_n.P_n is equal to,"},{"Start":"00:54.365 ","End":"00:56.690","Text":"just expand the second P_n,"},{"Start":"00:56.690 ","End":"00:58.580","Text":"leave the first one alone,"},{"Start":"00:58.580 ","End":"01:02.915","Text":"and it\u0027s a polynomial of degree n and n is bigger than 0,"},{"Start":"01:02.915 ","End":"01:06.475","Text":"and then we can use the linearity"},{"Start":"01:06.475 ","End":"01:10.359","Text":"to break it up into a sum and also to bring the constants in front."},{"Start":"01:10.359 ","End":"01:15.130","Text":"But remember they\u0027re conjugated because it\u0027s the second component of the 2."},{"Start":"01:15.130 ","End":"01:17.020","Text":"All these come out 0,"},{"Start":"01:17.020 ","End":"01:18.730","Text":"the ones that I\u0027ve put in gray,"},{"Start":"01:18.730 ","End":"01:20.990","Text":"we\u0027re just left with this one,"},{"Start":"01:20.990 ","End":"01:27.305","Text":"so we\u0027ve got that P_n with P_n is a_n conjugate P_n, x^n."},{"Start":"01:27.305 ","End":"01:29.290","Text":"Now towards the end of the previous clip,"},{"Start":"01:29.290 ","End":"01:34.320","Text":"we proved that this is true for k between 0 and n,"},{"Start":"01:34.320 ","End":"01:38.470","Text":"and if we replace k by n or just substitute k equals n,"},{"Start":"01:38.470 ","End":"01:40.210","Text":"it\u0027s good for this range also,"},{"Start":"01:40.210 ","End":"01:41.620","Text":"we get the following."},{"Start":"01:41.620 ","End":"01:44.420","Text":"It\u0027s pretty much this, but with n instead of k,"},{"Start":"01:44.420 ","End":"01:49.045","Text":"n minus n is 0, the 0th derivative is the function itself."},{"Start":"01:49.045 ","End":"01:51.815","Text":"Here it\u0027s just x^2 minus 1^n,"},{"Start":"01:51.815 ","End":"01:54.599","Text":"and we\u0027re going to do an integration by parts here,"},{"Start":"01:54.599 ","End":"01:56.780","Text":"but first let\u0027s split it up into 2 pieces,"},{"Start":"01:56.780 ","End":"01:59.030","Text":"x^2 minus 1 is x minus 1,"},{"Start":"01:59.030 ","End":"02:01.875","Text":"x plus 1 doesn\u0027t really matter, which is f\u0027,"},{"Start":"02:01.875 ","End":"02:03.440","Text":"which is g, but you have to choose,"},{"Start":"02:03.440 ","End":"02:05.660","Text":"so I\u0027ll choose this one is f\u0027(x),"},{"Start":"02:05.660 ","End":"02:07.520","Text":"so this one is g(x)."},{"Start":"02:07.520 ","End":"02:11.690","Text":"Using this formula, which you\u0027ve seen many times already in this course,"},{"Start":"02:11.690 ","End":"02:15.735","Text":"we get f is the integral of this,"},{"Start":"02:15.735 ","End":"02:18.634","Text":"g is this untouched,"},{"Start":"02:18.634 ","End":"02:20.870","Text":"here again, is f,"},{"Start":"02:20.870 ","End":"02:26.648","Text":"and then g\u0027 is derivative of this,"},{"Start":"02:26.648 ","End":"02:29.180","Text":"put an n in front and reduce the power by 1,"},{"Start":"02:29.180 ","End":"02:31.115","Text":"the internal derivative is 1."},{"Start":"02:31.115 ","End":"02:33.830","Text":"In this expression, if we plug in 1,"},{"Start":"02:33.830 ","End":"02:36.140","Text":"we get 0 because n is bigger than 0,"},{"Start":"02:36.140 ","End":"02:37.220","Text":"so it\u0027s n plus 1."},{"Start":"02:37.220 ","End":"02:40.190","Text":"If we plug in minus 1 also because this is bigger than 0,"},{"Start":"02:40.190 ","End":"02:41.375","Text":"we get a 0,"},{"Start":"02:41.375 ","End":"02:43.265","Text":"so this thing drops out,"},{"Start":"02:43.265 ","End":"02:48.755","Text":"and what we\u0027re left with is this constant times minus this integral."},{"Start":"02:48.755 ","End":"02:55.175","Text":"Bring the minus in front and then raise the power by 1,"},{"Start":"02:55.175 ","End":"02:58.850","Text":"and then combine n with n plus 1,"},{"Start":"02:58.850 ","End":"03:01.010","Text":"I mean as a fraction in front,"},{"Start":"03:01.010 ","End":"03:03.545","Text":"and what we\u0027re left with is this integral,"},{"Start":"03:03.545 ","End":"03:06.610","Text":"where here it\u0027s an n plus 1 here it\u0027s an n minus 1."},{"Start":"03:06.610 ","End":"03:09.630","Text":"Summarizing, this line is important, P_n, x^n,"},{"Start":"03:09.630 ","End":"03:11.970","Text":"and this one also P_n x^n,"},{"Start":"03:11.970 ","End":"03:16.055","Text":"and we develop this into this. New page."},{"Start":"03:16.055 ","End":"03:19.910","Text":"Now we can repeat all this integration by parts again."},{"Start":"03:19.910 ","End":"03:23.530","Text":"What we\u0027ll get is instead of a 1 here,"},{"Start":"03:23.530 ","End":"03:25.550","Text":"we\u0027ll have a 2 here."},{"Start":"03:25.550 ","End":"03:30.710","Text":"What happens in front is we just multiply by a new value of n,"},{"Start":"03:30.710 ","End":"03:32.540","Text":"n minus 1 over n plus 2."},{"Start":"03:32.540 ","End":"03:35.005","Text":"This keeps getting bigger by 1,"},{"Start":"03:35.005 ","End":"03:37.170","Text":"and this gets smaller by 1,"},{"Start":"03:37.170 ","End":"03:40.430","Text":"so the next time around we have this exponent from"},{"Start":"03:40.430 ","End":"03:44.165","Text":"here and this exponent from adding 1 to this one,"},{"Start":"03:44.165 ","End":"03:45.920","Text":"and we get x minus 1^n plus 2,"},{"Start":"03:45.920 ","End":"03:48.280","Text":"x plus 1^n minus 2."},{"Start":"03:48.280 ","End":"03:50.300","Text":"I think the pattern is clear,"},{"Start":"03:50.300 ","End":"03:54.770","Text":"so we\u0027re just going to continue and so on and so on really need induction."},{"Start":"03:54.770 ","End":"03:58.910","Text":"What we get if we put n here instead of the 1 or the 2,"},{"Start":"03:58.910 ","End":"04:01.130","Text":"we get this expression,"},{"Start":"04:01.130 ","End":"04:05.480","Text":"and here\u0027s what we get after simplifying n plus n is 2n, which is even,"},{"Start":"04:05.480 ","End":"04:07.802","Text":"so minus 1 to even is 1,"},{"Start":"04:07.802 ","End":"04:13.235","Text":"and then n times n minus 1 and so on down to 1 is n factorial."},{"Start":"04:13.235 ","End":"04:17.000","Text":"Here, we have going up from n plus 1 up to 2n."},{"Start":"04:17.000 ","End":"04:19.950","Text":"This drops off is the power of 0,"},{"Start":"04:19.950 ","End":"04:21.120","Text":"n plus n is 2n,"},{"Start":"04:21.120 ","End":"04:23.750","Text":"so this is what we have now, and this is good."},{"Start":"04:23.750 ","End":"04:27.445","Text":"We no longer have the 2 pieces with the x minus 1 and the x plus 1,"},{"Start":"04:27.445 ","End":"04:29.930","Text":"that\u0027s easier to do the integral."},{"Start":"04:29.930 ","End":"04:34.175","Text":"What we get is the integral of this,"},{"Start":"04:34.175 ","End":"04:38.263","Text":"just add 1 to the exponent and divide by that exponent,"},{"Start":"04:38.263 ","End":"04:41.950","Text":"then we have to substitute minus 1 to 1."},{"Start":"04:41.950 ","End":"04:43.170","Text":"In other words, plug in 1,"},{"Start":"04:43.170 ","End":"04:44.570","Text":"plug in minus 1 and subtract."},{"Start":"04:44.570 ","End":"04:48.650","Text":"What we get is if we put in the 1, we get 0,"},{"Start":"04:48.650 ","End":"04:49.910","Text":"put in the minus 1,"},{"Start":"04:49.910 ","End":"04:51.815","Text":"we get minus 2,"},{"Start":"04:51.815 ","End":"04:55.155","Text":"so here\u0027s the expression we have now."},{"Start":"04:55.155 ","End":"05:00.080","Text":"We have that P_n, x^n is equal to this,"},{"Start":"05:00.080 ","End":"05:04.805","Text":"equal to this, and it\u0027s now after simplification again, we have this."},{"Start":"05:04.805 ","End":"05:09.995","Text":"What I\u0027ve done here is I combined the minus to an odd number is minus,"},{"Start":"05:09.995 ","End":"05:11.690","Text":"and together with this minus,"},{"Start":"05:11.690 ","End":"05:13.640","Text":"the 2 minuses just disappear."},{"Start":"05:13.640 ","End":"05:15.170","Text":"The rest of it\u0027s the same."},{"Start":"05:15.170 ","End":"05:20.420","Text":"Now recall that P_n(x) is equal to this that was the definition."},{"Start":"05:20.420 ","End":"05:27.515","Text":"Let\u0027s expand x^2 minus 1^n using Newton\u0027s binomial expansion."},{"Start":"05:27.515 ","End":"05:29.540","Text":"If you don\u0027t remember, you should look it up by"},{"Start":"05:29.540 ","End":"05:34.615","Text":"Newton\u0027s binomial expansion in any web search you could find it probably."},{"Start":"05:34.615 ","End":"05:38.320","Text":"Now, note that the power is increased with k,"},{"Start":"05:38.320 ","End":"05:42.200","Text":"so the highest power will be when k is equal to n,"},{"Start":"05:42.200 ","End":"05:44.720","Text":"and then we\u0027ll get x^2n."},{"Start":"05:44.720 ","End":"05:50.220","Text":"All the rest of it will be monomials which together comprise a polynomial,"},{"Start":"05:50.220 ","End":"05:52.855","Text":"and the degree will be less than 2n."},{"Start":"05:52.855 ","End":"05:57.740","Text":"In fact, it\u0027s even less than or equal to 2n minus 2 for this P(x)."},{"Start":"05:57.740 ","End":"06:03.260","Text":"If we differentiate it n times the derivative of x^2n,"},{"Start":"06:03.260 ","End":"06:05.540","Text":"and times each time we get"},{"Start":"06:05.540 ","End":"06:09.860","Text":"the exponent put in front and reduce the exponent by 1 so that we get"},{"Start":"06:09.860 ","End":"06:17.150","Text":"2n and 2n minus 1 up to n plus 1 and then reducing it by 1 n times is x^n."},{"Start":"06:17.150 ","End":"06:22.388","Text":"As for this, is the nth derivative of a polynomial of degree 2n,"},{"Start":"06:22.388 ","End":"06:29.555","Text":"so its degree is less than n. This is a polynomial of degree n,"},{"Start":"06:29.555 ","End":"06:32.270","Text":"and this is the leading coefficient."},{"Start":"06:32.270 ","End":"06:36.635","Text":"What we\u0027ve expanded is this part."},{"Start":"06:36.635 ","End":"06:40.670","Text":"We still have this constant for P_n(x)."},{"Start":"06:40.670 ","End":"06:44.960","Text":"P_n(x) is that constant times what we have up here,"},{"Start":"06:44.960 ","End":"06:49.210","Text":"and the constant changes the polynomial but not the degree,"},{"Start":"06:49.210 ","End":"06:53.240","Text":"so we get a different polynomial of degree less than n. On the other hand,"},{"Start":"06:53.240 ","End":"06:57.830","Text":"we had that P_n equals a_n x^n plus stuff of degree"},{"Start":"06:57.830 ","End":"07:03.110","Text":"lower than n and so a_n is equal to this."},{"Start":"07:03.110 ","End":"07:05.390","Text":"We can simplify this, you don\u0027t have to."},{"Start":"07:05.390 ","End":"07:07.389","Text":"We could complete it here,"},{"Start":"07:07.389 ","End":"07:10.350","Text":"keep going down to 1 if we put n,"},{"Start":"07:10.350 ","End":"07:13.070","Text":"n minus 1 and so on so we get 2n factorial,"},{"Start":"07:13.070 ","End":"07:16.205","Text":"and we compensate by putting another n factorial down here."},{"Start":"07:16.205 ","End":"07:18.020","Text":"Optional to do this or not,"},{"Start":"07:18.020 ","End":"07:19.540","Text":"if you think it\u0027s neater."},{"Start":"07:19.540 ","End":"07:22.775","Text":"Now here are some formulas that we\u0027ve proved so far,"},{"Start":"07:22.775 ","End":"07:24.770","Text":"summarizing the essential bits."},{"Start":"07:24.770 ","End":"07:32.595","Text":"We just did a_n equals this and we had in a product of P_n with x_n here,"},{"Start":"07:32.595 ","End":"07:34.865","Text":"and P_n with P_n here."},{"Start":"07:34.865 ","End":"07:42.205","Text":"What we can do is take the a_n bar from here and the P_n, x^n from here."},{"Start":"07:42.205 ","End":"07:45.180","Text":"If we stick those together,"},{"Start":"07:45.180 ","End":"07:52.000","Text":"then what we get is this bit here times this bit here."},{"Start":"07:52.490 ","End":"07:57.818","Text":"I\u0027ve used colors you\u0027ll see and I see the stuff I could cancel."},{"Start":"07:57.818 ","End":"08:03.500","Text":"This 2^n, 2^n cancels with 2^2n and leaves us with a 2 here."},{"Start":"08:03.500 ","End":"08:06.455","Text":"This n factorial, with this n factorial,"},{"Start":"08:06.455 ","End":"08:08.720","Text":"this is the same as this,"},{"Start":"08:08.720 ","End":"08:10.540","Text":"just in reverse order."},{"Start":"08:10.540 ","End":"08:15.560","Text":"All we\u0027re left with at this point is a 2 from here,"},{"Start":"08:15.560 ","End":"08:18.560","Text":"and a 2n plus 1 here,"},{"Start":"08:18.560 ","End":"08:21.050","Text":"and that\u0027s what we had to prove,"},{"Start":"08:21.050 ","End":"08:23.940","Text":"and we\u0027re finally done."}],"ID":28708},{"Watched":false,"Name":"Exercise 10a - Laguerre Polynomials","Duration":"7m 5s","ChapterTopicVideoID":27475,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.282","Text":"Here\u0027s another exercise with polynomials,"},{"Start":"00:03.282 ","End":"00:05.730","Text":"this time Laguerre polynomials."},{"Start":"00:05.730 ","End":"00:10.270","Text":"K is the space of piece-wise continuous functions on the reals,"},{"Start":"00:10.270 ","End":"00:12.750","Text":"which satisfy this condition."},{"Start":"00:12.750 ","End":"00:14.880","Text":"There\u0027s an inner product like this."},{"Start":"00:14.880 ","End":"00:16.755","Text":"This is the weight function."},{"Start":"00:16.755 ","End":"00:23.220","Text":"This is actually the norm squared that\u0027s induced from the inner product."},{"Start":"00:23.220 ","End":"00:26.770","Text":"Polynomials L_n, L for Laguerre,"},{"Start":"00:26.770 ","End":"00:29.610","Text":"defined as 1 over n factorial e^x,"},{"Start":"00:29.610 ","End":"00:31.425","Text":"nth derivative of x^n,"},{"Start":"00:31.425 ","End":"00:33.360","Text":"e to the minus x."},{"Start":"00:33.360 ","End":"00:37.110","Text":"We have to prove that this is an orthonormal system,"},{"Start":"00:37.110 ","End":"00:40.245","Text":"not just orthogonal like with the other polynomials."},{"Start":"00:40.245 ","End":"00:41.525","Text":"We had Legendre,"},{"Start":"00:41.525 ","End":"00:43.160","Text":"we had Chebyshev,"},{"Start":"00:43.160 ","End":"00:45.602","Text":"and we had Hermite. They were orthogonal."},{"Start":"00:45.602 ","End":"00:47.908","Text":"This one is also orthonormal,"},{"Start":"00:47.908 ","End":"00:51.110","Text":"and we\u0027re given a useful formula, this integral."},{"Start":"00:51.110 ","End":"00:53.030","Text":"Let\u0027s start solving it."},{"Start":"00:53.030 ","End":"00:56.720","Text":"L_n is a degree and polynomial."},{"Start":"00:56.720 ","End":"01:01.025","Text":"Here\u0027s a list of the first few Laguerre polynomials."},{"Start":"01:01.025 ","End":"01:02.390","Text":"I won\u0027t prove it formally,"},{"Start":"01:02.390 ","End":"01:05.124","Text":"but I\u0027ll give you a quick proof."},{"Start":"01:05.124 ","End":"01:09.385","Text":"If we take an nth degree polynomial, instead of this,"},{"Start":"01:09.385 ","End":"01:11.473","Text":"times e to the minus x,"},{"Start":"01:11.473 ","End":"01:13.310","Text":"and we differentiate it,"},{"Start":"01:13.310 ","End":"01:17.120","Text":"we also get a degree n polynomial times e to the minus x."},{"Start":"01:17.120 ","End":"01:18.650","Text":"That\u0027s fairly easy to see."},{"Start":"01:18.650 ","End":"01:20.654","Text":"No matter how many times you do it,"},{"Start":"01:20.654 ","End":"01:23.890","Text":"we\u0027ll get an nth degree polynomial times e to the minus x."},{"Start":"01:23.890 ","End":"01:25.460","Text":"That\u0027s this part."},{"Start":"01:25.460 ","End":"01:26.840","Text":"If you differentiate it n times,"},{"Start":"01:26.840 ","End":"01:29.115","Text":"it doesn\u0027t matter how many times, that\u0027s what we\u0027ll get."},{"Start":"01:29.115 ","End":"01:30.875","Text":"If you multiply by e^x,"},{"Start":"01:30.875 ","End":"01:34.028","Text":"that will cancel out with the e to the minus x that remains,"},{"Start":"01:34.028 ","End":"01:36.380","Text":"and so we\u0027ll just get the nth degree polynomial."},{"Start":"01:36.380 ","End":"01:37.985","Text":"That was an informal proof."},{"Start":"01:37.985 ","End":"01:41.930","Text":"Now, we need to show orthogonality, first of all,"},{"Start":"01:41.930 ","End":"01:46.760","Text":"that L_n in a product with L_m is 0 if n is not equal to m. Later,"},{"Start":"01:46.760 ","End":"01:48.590","Text":"we\u0027ll do the orthonormality."},{"Start":"01:48.590 ","End":"01:50.600","Text":"We had a previous exercise,"},{"Start":"01:50.600 ","End":"01:52.865","Text":"and I\u0027ll show it to you again."},{"Start":"01:52.865 ","End":"01:55.190","Text":"This exercise, which we proved,"},{"Start":"01:55.190 ","End":"02:01.320","Text":"shows that we don\u0027t have to show orthogonality of a set."}],"ID":28709},{"Watched":false,"Name":"Exercise 10b - Laguerre Polynomials","Duration":"4m 2s","ChapterTopicVideoID":27476,"CourseChapterTopicPlaylistID":274168,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:02.670","Text":"Back after the break,"},{"Start":"00:02.670 ","End":"00:09.495","Text":"what remains to show is that the norm of L_n is 1 or norm of L_n^2,"},{"Start":"00:09.495 ","End":"00:13.575","Text":"which is the inner product of L_n with itself."},{"Start":"00:13.575 ","End":"00:16.755","Text":"Just to remind you what the exercise is,"},{"Start":"00:16.755 ","End":"00:20.505","Text":"we\u0027re continuing with the Laguerre polynomials."},{"Start":"00:20.505 ","End":"00:26.925","Text":"Let\u0027s replace the second L_n by the polynomial that it is."},{"Start":"00:26.925 ","End":"00:31.665","Text":"Then using linearity, we get this sum."},{"Start":"00:31.665 ","End":"00:33.840","Text":"When we take out the scalars,"},{"Start":"00:33.840 ","End":"00:36.945","Text":"they come out conjugated because it\u0027s the second parameter."},{"Start":"00:36.945 ","End":"00:41.790","Text":"Now we already showed that L_n with x^k is 0 for"},{"Start":"00:41.790 ","End":"00:47.000","Text":"all k less than n. So we\u0027re just left with the first term."},{"Start":"00:47.000 ","End":"00:54.940","Text":"We have that the norm of L_n^2 is a_n conjugate times L_n dot x^n."},{"Start":"00:54.940 ","End":"00:58.880","Text":"Now earlier we showed that this expression here is"},{"Start":"00:58.880 ","End":"01:03.495","Text":"equal to this and it\u0027s good also for k=n."},{"Start":"01:03.495 ","End":"01:10.190","Text":"Putting k=n here and here and here, we get this."},{"Start":"01:10.190 ","End":"01:12.360","Text":"Now, n minus n is 0."},{"Start":"01:12.360 ","End":"01:15.800","Text":"The 0th derivative is the function itself."},{"Start":"01:15.800 ","End":"01:19.550","Text":"What we get is the integral of x^n,"},{"Start":"01:19.550 ","End":"01:22.825","Text":"e^x from 0 to infinity."},{"Start":"01:22.825 ","End":"01:27.050","Text":"This integral is equal to n factorial."},{"Start":"01:27.050 ","End":"01:29.960","Text":"Note that if p(x) is a polynomial,"},{"Start":"01:29.960 ","End":"01:34.520","Text":"which is x^n plus terms of degree less than n,"},{"Start":"01:34.520 ","End":"01:37.333","Text":"all this times e^minus x,"},{"Start":"01:37.333 ","End":"01:39.350","Text":"then when you differentiate it,"},{"Start":"01:39.350 ","End":"01:44.390","Text":"you get the derivative of e^minus x times this,"},{"Start":"01:44.390 ","End":"01:51.060","Text":"which is minus e^minus x times x^n plus lower power."},{"Start":"01:51.060 ","End":"01:54.460","Text":"Derivative of this as the lower power."},{"Start":"01:55.940 ","End":"01:59.930","Text":"Basically, the derivative of p is minus"},{"Start":"01:59.930 ","End":"02:03.620","Text":"x^n plus a polynomial of degree less than n also,"},{"Start":"02:03.620 ","End":"02:05.080","Text":"not the same one."},{"Start":"02:05.080 ","End":"02:10.310","Text":"The only thing that essentially changes is the sign in front of the x^n."},{"Start":"02:10.310 ","End":"02:13.055","Text":"We don\u0027t care what kind of polynomial this is,"},{"Start":"02:13.055 ","End":"02:17.195","Text":"except that it\u0027s of degree less than n. By induction,"},{"Start":"02:17.195 ","End":"02:20.239","Text":"the kth derivative is minus 1^k x^n"},{"Start":"02:20.239 ","End":"02:23.780","Text":"plus yet another polynomial of degree less than n. Well,"},{"Start":"02:23.780 ","End":"02:26.290","Text":"this times e^minus x."},{"Start":"02:26.290 ","End":"02:29.170","Text":"Now L_n of x,"},{"Start":"02:29.170 ","End":"02:32.075","Text":"by definition is equal to this."},{"Start":"02:32.075 ","End":"02:36.980","Text":"By the above, this is equal to minus 1^n,"},{"Start":"02:36.980 ","End":"02:40.955","Text":"x^n plus polynomial degree less than n, e^minus x."},{"Start":"02:40.955 ","End":"02:47.240","Text":"That\u0027s the last part and we have 1 over n factorial e^x."},{"Start":"02:47.240 ","End":"02:53.530","Text":"E^x cancels with e^minus x. E^x cancels with e^minus x."},{"Start":"02:53.530 ","End":"02:59.990","Text":"The 1 over n factorial multiply it and the polynomial of degree less than n times"},{"Start":"02:59.990 ","End":"03:02.570","Text":"the constant is still a polynomial of degree less than"},{"Start":"03:02.570 ","End":"03:06.570","Text":"n. This is what we have for L_n(x)."},{"Start":"03:06.570 ","End":"03:12.695","Text":"On the other hand, this is equal to a_n x^n plus a_n minus 1 x^n minus 1."},{"Start":"03:12.695 ","End":"03:16.820","Text":"All this is a polynomial of degree less than n. We can conclude from"},{"Start":"03:16.820 ","End":"03:21.140","Text":"this that the coefficient of x^n here equals the coefficient of x^n here."},{"Start":"03:21.140 ","End":"03:27.445","Text":"In other words that a_n is equal to minus 1^n over n factorial."},{"Start":"03:27.445 ","End":"03:34.080","Text":"But earlier we showed this and we also showed that this part here,"},{"Start":"03:34.080 ","End":"03:37.325","Text":"L_n, x^n is minus 1^n, n factorial."},{"Start":"03:37.325 ","End":"03:39.080","Text":"We combine all this,"},{"Start":"03:39.080 ","End":"03:44.160","Text":"we get that this is equal the a_n-bar from"},{"Start":"03:44.160 ","End":"03:50.295","Text":"here is this and the second part is equal to this."},{"Start":"03:50.295 ","End":"03:57.090","Text":"Everything cancels, n factorial with n factorial minus 1^2n is 1."},{"Start":"03:57.090 ","End":"03:59.160","Text":"We just get 1."},{"Start":"03:59.160 ","End":"04:03.490","Text":"That\u0027s what we had to show and we are done."}],"ID":28710}],"Thumbnail":null,"ID":274168},{"Name":"Summarizing Exercises","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1 - Part a","Duration":"4m 51s","ChapterTopicVideoID":27443,"CourseChapterTopicPlaylistID":274169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/27443.jpeg","UploadDate":"2021-12-01T10:22:19.3100000","DurationForVideoObject":"PT4M51S","Description":null,"MetaTitle":"Exercise 1 - Part a - Summarizing Exercises: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Summarizing Exercises practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/calculus-i%2c-ii-and-iii/inner-product-spaces%2c-normed-spaces/summarizing-exercises/vid28662","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"In this exercise, V is the vector space of"},{"Start":"00:03.390 ","End":"00:07.590","Text":"piecewise continuous functions from the interval minus Pi,"},{"Start":"00:07.590 ","End":"00:11.324","Text":"Pi to the complex numbers."},{"Start":"00:11.324 ","End":"00:16.965","Text":"It has an inner product given as follows."},{"Start":"00:16.965 ","End":"00:20.295","Text":"We don\u0027t have to prove that this is an inner product."},{"Start":"00:20.295 ","End":"00:25.245","Text":"In part a, we have to prove that the set e to the inx,"},{"Start":"00:25.245 ","End":"00:29.970","Text":"where n is an integer from minus infinity to infinity."},{"Start":"00:29.970 ","End":"00:37.965","Text":"This set is an orthogonal system on V. We have to find the norm of e to the inx."},{"Start":"00:37.965 ","End":"00:39.960","Text":"It would be 1 if it was orthonormal."},{"Start":"00:39.960 ","End":"00:42.285","Text":"That\u0027s why it makes sense to ask for it."},{"Start":"00:42.285 ","End":"00:45.860","Text":"The norm is the 1 induced by the inner product."},{"Start":"00:45.860 ","End":"00:49.280","Text":"Part b, we\u0027ll read it out when we come to it."},{"Start":"00:49.280 ","End":"00:51.700","Text":"Let\u0027s start with part a."},{"Start":"00:51.700 ","End":"00:56.705","Text":"The idea in part a is to take the inner product of 2 of these,"},{"Start":"00:56.705 ","End":"00:57.875","Text":"e to the inx,"},{"Start":"00:57.875 ","End":"00:59.840","Text":"e to the imx."},{"Start":"00:59.840 ","End":"01:04.880","Text":"After developing it, we should get that this is equal to 0 if and"},{"Start":"01:04.880 ","End":"01:09.565","Text":"only if n is not equal to m. By the definition,"},{"Start":"01:09.565 ","End":"01:13.400","Text":"here it is, this will be the inner product f,"},{"Start":"01:13.400 ","End":"01:16.760","Text":"g here, and f\u0027 g\u0027."},{"Start":"01:16.760 ","End":"01:19.310","Text":"Derivative of e to the inx is in,"},{"Start":"01:19.310 ","End":"01:20.645","Text":"e to the inx."},{"Start":"01:20.645 ","End":"01:26.630","Text":"Now the conjugate of e to the imx is e to the minus imx."},{"Start":"01:26.630 ","End":"01:33.880","Text":"I\u0027ll remind you why this represents a point on the unit circle with angle mx radians."},{"Start":"01:33.880 ","End":"01:36.560","Text":"If you reflected in the x-axis,"},{"Start":"01:36.560 ","End":"01:39.410","Text":"we need minus mx radians,"},{"Start":"01:39.410 ","End":"01:43.640","Text":"just instead of going counterclockwise, we go clockwise."},{"Start":"01:43.640 ","End":"01:47.240","Text":"Here also, the conjugate of this, like we said,"},{"Start":"01:47.240 ","End":"01:51.725","Text":"is this, and the conjugate of im is minus im."},{"Start":"01:51.725 ","End":"01:54.290","Text":"The conjugate of a product is the product of the conjugates."},{"Start":"01:54.290 ","End":"02:00.965","Text":"This is what we get for f\u0027(x) times g\u0027(x)."},{"Start":"02:00.965 ","End":"02:03.260","Text":"Now we can simplify a bit."},{"Start":"02:03.260 ","End":"02:08.695","Text":"This exponent can be combined with the rules of exponents. Same here."},{"Start":"02:08.695 ","End":"02:17.065","Text":"We can also bring the in with the minus im up in front to get minus i^2 nm."},{"Start":"02:17.065 ","End":"02:18.950","Text":"Now the integrals are the same,"},{"Start":"02:18.950 ","End":"02:22.795","Text":"so we just have to combine the coefficient in front of them."},{"Start":"02:22.795 ","End":"02:25.875","Text":"Remember that minus i^2 is plus 1."},{"Start":"02:25.875 ","End":"02:29.355","Text":"We have 1 over Pi plus nm over Pi,"},{"Start":"02:29.355 ","End":"02:34.080","Text":"which means 1 plus nm over Pi times this integral."},{"Start":"02:34.080 ","End":"02:38.510","Text":"To do this integral, we need to distinguish 2 cases;"},{"Start":"02:38.510 ","End":"02:44.320","Text":"n equals m or n not equals m. Because if n equals m,"},{"Start":"02:44.320 ","End":"02:46.450","Text":"this is not an exponent anymore."},{"Start":"02:46.450 ","End":"02:48.970","Text":"We\u0027ll do both cases."},{"Start":"02:48.970 ","End":"02:51.640","Text":"If n is not equal to m,"},{"Start":"02:51.640 ","End":"02:57.520","Text":"then we get that this inner product is 1 plus nm over Pi."},{"Start":"02:57.520 ","End":"02:59.260","Text":"I\u0027m just copying this."},{"Start":"02:59.260 ","End":"03:04.920","Text":"Now, the integral of this is e to the in minus mx."},{"Start":"03:04.920 ","End":"03:07.750","Text":"We divide by the coefficient of x,"},{"Start":"03:07.750 ","End":"03:14.135","Text":"which is in minus m. Then we evaluate it between minus Pi and Pi."},{"Start":"03:14.135 ","End":"03:17.404","Text":"This is equal to the following,"},{"Start":"03:17.404 ","End":"03:22.130","Text":"because e to the i Pi is minus 1."},{"Start":"03:22.130 ","End":"03:26.245","Text":"And e to the minus i Pi is minus 1."},{"Start":"03:26.245 ","End":"03:30.500","Text":"We still have to the power of n minus m. The point is,"},{"Start":"03:30.500 ","End":"03:32.810","Text":"it doesn\u0027t matter if n minus m is odd or even."},{"Start":"03:32.810 ","End":"03:34.220","Text":"It\u0027s the same thing here and here."},{"Start":"03:34.220 ","End":"03:37.345","Text":"We\u0027ve got something minus itself, so that\u0027s 0."},{"Start":"03:37.345 ","End":"03:39.275","Text":"When n is not equal to m,"},{"Start":"03:39.275 ","End":"03:41.825","Text":"this inner product is 0, and that\u0027s good."},{"Start":"03:41.825 ","End":"03:45.770","Text":"Let\u0027s take the other case where n equals m. In that case,"},{"Start":"03:45.770 ","End":"03:52.445","Text":"we\u0027ll just replace m by n. We get 1 plus n^2 over Pi."},{"Start":"03:52.445 ","End":"03:55.730","Text":"Then the integral n minus m is 0."},{"Start":"03:55.730 ","End":"03:58.285","Text":"This is e to the 0 which is 1."},{"Start":"03:58.285 ","End":"04:03.290","Text":"This is just equal to 1 plus n^2 over Pi times the length of the interval,"},{"Start":"04:03.290 ","End":"04:07.025","Text":"which is 2Pi, Pi cancels with Pi,"},{"Start":"04:07.025 ","End":"04:10.230","Text":"and we get twice 1 plus n^2."},{"Start":"04:10.230 ","End":"04:17.880","Text":"We\u0027ve shown that this is an orthogonal system because of the 0 here and non-zero here."},{"Start":"04:17.880 ","End":"04:21.965","Text":"We were also asked to find the norm of e to the inx."},{"Start":"04:21.965 ","End":"04:28.280","Text":"Notice that we can replace this m by n. This inner product is the norm squared."},{"Start":"04:28.280 ","End":"04:30.100","Text":"We are talking about the induced norm."},{"Start":"04:30.100 ","End":"04:32.000","Text":"That\u0027s the definition of the induced norm."},{"Start":"04:32.000 ","End":"04:34.340","Text":"If the norm squared is equal to this,"},{"Start":"04:34.340 ","End":"04:37.915","Text":"then the norm is equal to the square root of this."},{"Start":"04:37.915 ","End":"04:46.760","Text":"We\u0027ve answered both parts of part a orthogonality and the norm of e to the inx."},{"Start":"04:46.760 ","End":"04:51.750","Text":"We\u0027ll do part b in the following clip."}],"ID":28662},{"Watched":false,"Name":"Exercise 1 - Part b","Duration":"6m 27s","ChapterTopicVideoID":27444,"CourseChapterTopicPlaylistID":274169,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.170","Text":"We just finished part a of this exercise where we"},{"Start":"00:04.170 ","End":"00:07.995","Text":"showed that this is an orthogonal system."},{"Start":"00:07.995 ","End":"00:16.380","Text":"We also found that this norm is equal to square root of twice 1 plus n^2."},{"Start":"00:16.380 ","End":"00:20.610","Text":"We have to prove that there\u0027s no such f in v,"},{"Start":"00:20.610 ","End":"00:22.560","Text":"such that this limit,"},{"Start":"00:22.560 ","End":"00:26.880","Text":"as n goes to infinity of this expression is equal to 1."},{"Start":"00:26.880 ","End":"00:28.515","Text":"Although this looks a mess,"},{"Start":"00:28.515 ","End":"00:31.665","Text":"we\u0027ll do some tidying up and you\u0027ll see it\u0027s not so bad."},{"Start":"00:31.665 ","End":"00:34.920","Text":"Anyway, we\u0027re going to do the proof by contradiction."},{"Start":"00:34.920 ","End":"00:39.300","Text":"We\u0027re going to assume that such an f does exist,"},{"Start":"00:39.300 ","End":"00:41.670","Text":"that has this property,"},{"Start":"00:41.670 ","End":"00:44.600","Text":"and we\u0027ll reach a contradiction."},{"Start":"00:44.600 ","End":"00:45.920","Text":"So we start off with this,"},{"Start":"00:45.920 ","End":"00:48.545","Text":"at this limit is equal to 1,"},{"Start":"00:48.545 ","End":"00:50.825","Text":"and then we\u0027ll just work on this."},{"Start":"00:50.825 ","End":"00:55.820","Text":"We can put e to the minus nx inside the brackets here,"},{"Start":"00:55.820 ","End":"01:02.570","Text":"and also put the minus in after the f\u0027."},{"Start":"01:02.570 ","End":"01:09.320","Text":"Notice that the derivative of e to the minus inx is minus in,"},{"Start":"01:09.320 ","End":"01:12.265","Text":"e to the minus inx."},{"Start":"01:12.265 ","End":"01:21.305","Text":"This f\u0027 is the derivative of f. This very much reminds us of the inner product,"},{"Start":"01:21.305 ","End":"01:24.341","Text":"but we still need some conjugates."},{"Start":"01:24.341 ","End":"01:31.950","Text":"For that, we\u0027ll see that e to the minus inx is the conjugate of e to the inx."},{"Start":"01:32.300 ","End":"01:36.365","Text":"This is the conjugate of this."},{"Start":"01:36.365 ","End":"01:39.800","Text":"What we have here is almost an inner product except that there\u0027s"},{"Start":"01:39.800 ","End":"01:43.190","Text":"no 1 over Pi in front of the integral."},{"Start":"01:43.190 ","End":"01:46.280","Text":"So we can just put a Pi there,"},{"Start":"01:46.280 ","End":"01:47.960","Text":"and that\u0027ll make it all right."},{"Start":"01:47.960 ","End":"01:54.800","Text":"It\u0027s Pi times the inner product of f with the function e to the inx here."},{"Start":"01:54.800 ","End":"02:00.100","Text":"Now we can put a 2 at the top and at the bottom,"},{"Start":"02:00.100 ","End":"02:07.690","Text":"because this 2 times 1 plus n^2 is the norm squared of e to the inx we saw in part a."},{"Start":"02:07.690 ","End":"02:15.940","Text":"What we can do is to push this inside the absolute value squared."},{"Start":"02:15.940 ","End":"02:18.399","Text":"This is real and positive,"},{"Start":"02:18.399 ","End":"02:22.870","Text":"so we can put it in as the square root of"},{"Start":"02:22.870 ","End":"02:28.344","Text":"the same thing because it\u0027s going to come out right when it\u0027s squared."},{"Start":"02:28.344 ","End":"02:31.255","Text":"Now, like I said,"},{"Start":"02:31.255 ","End":"02:32.590","Text":"and we saw this in part a."},{"Start":"02:32.590 ","End":"02:37.375","Text":"This square root is equal to this norm."},{"Start":"02:37.375 ","End":"02:41.810","Text":"This has norm 1."},{"Start":"02:41.810 ","End":"02:46.943","Text":"If we replace e to the inx by e to the inx over its normal,"},{"Start":"02:46.943 ","End":"02:49.975","Text":"we\u0027ll get an orthonormal set."},{"Start":"02:49.975 ","End":"02:52.880","Text":"Call this e_n."},{"Start":"02:52.880 ","End":"02:58.410","Text":"So e_n of x is e to the inx over its norm,"},{"Start":"02:58.410 ","End":"03:04.235","Text":"so that e_n is now an orthonormal system, not merely orthogonal."},{"Start":"03:04.235 ","End":"03:09.655","Text":"Now if I replace this minus infinity by 1,"},{"Start":"03:09.655 ","End":"03:11.840","Text":"I have a smaller set,"},{"Start":"03:11.840 ","End":"03:14.000","Text":"but it\u0027s still orthonormal."},{"Start":"03:14.000 ","End":"03:17.360","Text":"The reason I want from 1 to infinity is that we\u0027re taking"},{"Start":"03:17.360 ","End":"03:22.130","Text":"the limit as n goes to plus infinity."},{"Start":"03:22.130 ","End":"03:25.010","Text":"We don\u0027t need the negative ends."},{"Start":"03:25.010 ","End":"03:31.440","Text":"You\u0027ll see copying from here is that this limit is equal to 1."},{"Start":"03:31.440 ","End":"03:37.160","Text":"Informally, we can say that this squared"},{"Start":"03:37.160 ","End":"03:43.790","Text":"is roughly 1 over 2nPi when n is very large,"},{"Start":"03:43.790 ","End":"03:48.240","Text":"like the quotient of these 2 goes to 1."},{"Start":"03:48.240 ","End":"03:50.550","Text":"This is of the order of this."},{"Start":"03:50.550 ","End":"03:52.310","Text":"We\u0027ll make it more precise in a moment."},{"Start":"03:52.310 ","End":"03:55.055","Text":"This is just to get an intuitive feel."},{"Start":"03:55.055 ","End":"03:57.005","Text":"If we take the square root,"},{"Start":"03:57.005 ","End":"04:00.740","Text":"we get that the absolute value of this inner product is"},{"Start":"04:00.740 ","End":"04:05.680","Text":"about 1 over square root of 2 and Pi."},{"Start":"04:05.680 ","End":"04:08.585","Text":"We can use Bessel\u0027s inequality."},{"Start":"04:08.585 ","End":"04:12.900","Text":"We\u0027ll just take the sum on one side infinity."},{"Start":"04:12.900 ","End":"04:14.885","Text":"We don\u0027t need from minus infinity."},{"Start":"04:14.885 ","End":"04:16.460","Text":"That\u0027s what I was saying earlier."},{"Start":"04:16.460 ","End":"04:18.500","Text":"We just need from 1 to infinity,"},{"Start":"04:18.500 ","End":"04:22.685","Text":"and this is going to be less than or equal to the norm of f squared,"},{"Start":"04:22.685 ","End":"04:28.485","Text":"so that this sum converges, it\u0027s finite."},{"Start":"04:28.485 ","End":"04:31.910","Text":"We\u0027re heading for a contradiction because I\u0027m going to show you that"},{"Start":"04:31.910 ","End":"04:35.660","Text":"on the other hand this diverges."},{"Start":"04:35.660 ","End":"04:40.895","Text":"Because this sum is of the order of,"},{"Start":"04:40.895 ","End":"04:43.620","Text":"again speaking informally,"},{"Start":"04:43.620 ","End":"04:46.605","Text":"sum of 1 over 2nPi,"},{"Start":"04:46.605 ","End":"04:49.275","Text":"and the 2Pi is a constant."},{"Start":"04:49.275 ","End":"04:52.010","Text":"It\u0027s like the sum of 1 over n,"},{"Start":"04:52.010 ","End":"04:53.930","Text":"which is the harmonic series,"},{"Start":"04:53.930 ","End":"04:55.925","Text":"and so it diverges,"},{"Start":"04:55.925 ","End":"04:58.490","Text":"and that\u0027s a contradiction."},{"Start":"04:58.490 ","End":"05:02.180","Text":"Now what I have to do is clean up the informal bit,"},{"Start":"05:02.180 ","End":"05:04.700","Text":"explain it a bit more rigorously,"},{"Start":"05:04.700 ","End":"05:07.165","Text":"then we\u0027ll be done."},{"Start":"05:07.165 ","End":"05:09.410","Text":"Yeah, being more precise,"},{"Start":"05:09.410 ","End":"05:11.960","Text":"we\u0027re going to use the limit comparison test,"},{"Start":"05:11.960 ","End":"05:16.984","Text":"which says that if we have two infinite series of positive numbers"},{"Start":"05:16.984 ","End":"05:22.810","Text":"such that the ratio of corresponding terms tends to a limit L,"},{"Start":"05:22.810 ","End":"05:26.580","Text":"and that L is a positive number,"},{"Start":"05:26.580 ","End":"05:28.995","Text":"is not 0 and it\u0027s not infinity,"},{"Start":"05:28.995 ","End":"05:31.505","Text":"then the two series,"},{"Start":"05:31.505 ","End":"05:35.060","Text":"they either both converge or both diverge."},{"Start":"05:35.060 ","End":"05:45.895","Text":"In our case, we can take a_n to be this and b_n to be 1 over 2nPi."},{"Start":"05:45.895 ","End":"05:54.000","Text":"We saw that the limit of a_n over b_n goes to 1, it\u0027s 1."},{"Start":"05:54.000 ","End":"05:58.009","Text":"Divide a_n over b_n and we get this."},{"Start":"05:58.009 ","End":"05:59.990","Text":"Yeah, there should be an extra Pi here,"},{"Start":"05:59.990 ","End":"06:01.955","Text":"it doesn\u0027t make much difference though."},{"Start":"06:01.955 ","End":"06:07.250","Text":"Now, since this sum of 1 over 2nPi is like the sum of 1"},{"Start":"06:07.250 ","End":"06:12.995","Text":"over n diverges by the limit comparison test, so does this."},{"Start":"06:12.995 ","End":"06:15.935","Text":"But like we saw before, that\u0027s a contradiction."},{"Start":"06:15.935 ","End":"06:18.410","Text":"We noticed that on the one hand it converges,"},{"Start":"06:18.410 ","End":"06:20.960","Text":"and then on the other hand it diverges."},{"Start":"06:20.960 ","End":"06:23.930","Text":"That concludes part b,"},{"Start":"06:23.930 ","End":"06:27.180","Text":"and that\u0027s the end of this exercise."}],"ID":28663}],"Thumbnail":null,"ID":274169}]

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1.1

1